Question 1

Question 1: Score 0/2We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 13 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.072. This means:

Incorrect

Your Answer:Correct Answer:

There was only a 7.2% chance of observing an increase greater than 13 kg (assuming the null hypothesis was true).

Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is true.

Question 2: Score 0/2

An agronomist measured the height of 150 Wheat plants. The mean height was 204 cm and the standard deviation was 17 cm. Calculate the standard error of the mean. Incorrect

Your Answer:Correct Answer: 1.388Comment:

The Standard Error is , where s is sample SD.

Notice that the sample mean does not matter for this result!


A medical researcher is interested in whether patients' left arms or right arms are longer. If 11 patients participate in this study (so that 11 left arms and 11 right arms are measured), how many degrees of freedom should the researcher use in her t-test critical value?

Incorrect


10

Comment: The attribute under study is the difference in arm length from a person's left to right arm. Since there is one observation of this per individual in the population, the degrees of freedom is 11 - 1 = 10


To determine the reliability of experts used in interpreting the results of polygraph examinations in criminal investigations, 300 cases were studied. The results were:

True Status

Innocent Guilty

Examiner'sDecision

Innocent 19 29

Guilty 131 121

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type I error as:

Incorrect


0.873

Comment:

Type I error is when a true null hypothesis is rejected. In this case, this is the probability of the Examiner judging an innocent suspect to be guilty. The probability is:


Researchers tested patients fitted with a medical telemetry unit to see if use of a cellular telephone interferes with the operation of the device. There were 555 tests conducted for one type of cellular telephone; interference with the device was found in 15% of these tests. Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p.

Incorrect

Your Answer:Correct Answer: (0.111, 0.189)Comment: The confidence interval would be:

=

= (0.111, 0.189)

https://maple-ta.uwaterloo.ca/mapleta/web/StaRooCo000/Public_Html/Tools/NormalCalculator.htm%00%E5%A1%B9%EF%92%81%E1%B4%BB%E4%A1%BF%E2%B2%AF%E5%B6%82%E8%97%84%E6%8C%A7%00%00%EA%AE%A5


In an stock portfolio selection process, a financial consultant observed the value of 45 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.53 (in millions) and the standard deviation is USD 0.19 (millions). Assuming normality, a 90% confidence interval is:

Incorrect


(1.483 , 1.577)

Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:


A random sample of 505 printers discovered that 68 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses. Incorrect

Your Answer:Correct Answer: 0.096 < p < 0.174Comment:


An agronomist measured the height of 132 Corn plants. The mean height was 239 cm and the standard deviation was 17 cm. Calculate the standard

error of the mean. IncorrectYour Answer:Correct Answer: 1.4797Comment:

The Standard Error is , where s is sample

SD.





A pharmacist is planning to estimate the mean level of a certain drug in a lab. The pharmacist wanted the estimate to be within 7 mg/dLi or less with 95% confidence. The pharmacist also believes that the standard deviation of the drug level is probably about 41 mg/dLi. How large a sample should the pharmacist need to take?

Incorrect

Your Answer:Correct Answer: 132Comment:

Use the equation: . For a 95% confidence interval we have c = 1.96,

so:



error of the mean. (3 decimal accuracy) IncorrectYour Answer:

Correct Answer: 1.1722±0.01Comment:


SD.



6 squirrels were found to have an average weight of 390 grams with a sample standard deviation of 5.35. Find the 95% confidence interval of the true mean weight (assume the t-student distribution).

Incorrect


(384.38,395.62)



Comment:Use the formula . With degrees of freedom = 5 we have t0.025 =

2.571 so our confidence interval is


We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 7.6 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.094. This means:

Incorrect


There was only a 9.4% chance of observing an increase greater than 7.6 kg (assuming the null hypothesis was true).

Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals (units) while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more different mean weight, so more weight gained) occurrence, given that the null hypothesis is true.


In order to study the harmful effects of DDT poisoning, the pesticide was fed to 6 randomly chosen rats out of a group of 12 rats. The other 6 rats were used as the control group. The following data gives the measurements of the amount of tremor detected in the bodies of each rat after the experiment: The more tremor, the more harmful.

Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2

A computer analysis is done with the output below (the differences are computed as control - poisoned)

Incorrect

t-test Difference t-test DF Prob>|t|

Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Upper 95% -2.1003

(Assuming equal variances.)

Which of the following is correct?Your Answer:Correct Answer:

The confidence interval does not include 0. Hence, there is evidence that the mean number of tremors for all potential rats in the poisoned group is larger than that in the control group.

Comment: The confidence interval shows the range that 95% of data should fall into given previous information. Since this range does not include 0, at least 95% of the time the poisoned rats will not have the same mean numbers of tremors as the control rats.


A study of 95 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 10 m. Which of the following is the 97% confidence interval for the mean length per bolt of carpet? Incorrect


(179.77, 184.23)

Comment: For a 97% CI we need to find the z value for the standard normal for which 98.5% of the graph area lies to the left. This is 2.1701 . The CI is:


In a sample of 530 mice, a biologist found that 65% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less. Incorrect

Your Answer:Correct Answer: 0.609% < p < 0.69%Comment:




In hypothesis testing, β is the probability of committing an error of Type II. The power of the test, 1 − β is then:

IncorrectYour Answer:Correct Answer:

the probability of rejecting H0 when HA is true

Comment: Type II error is when a false null hypothesis is not rejected. β is the probability of failing to reject null hypothesis H0 given that HA is true. 1-β is the opposite of this, so the probability of rejecting H0 given HA is true.


The Bata Shoe Museum states that 33% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.01 of the true proportion. How large a sample is necessary?

Incorrect

Your Answer:Correct Answer: 11,965Comment:


The formula of the t -test for dependent samples is:


Comment:


A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of small nests and the other of large nests. She measures one egg from each nest. The data are summarized below.

Incorrect


The null and alternate hypothesis of interest is:Your Answer:Correct Answer:

H : μL = μS; A : μL > μS

Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger

nests is larger than the mean egg size in smaller nests, or . The null

hypothesis is that the egg size is the same no matter which nest the egg came from

from, .


A survey of 519 women shoppers found that 38% of them shop on impulse. What is the 98% confidence interval for the true proportion of

women shoppers who shop on impulse? Incorrect


0.3304 < p < 0.4296

Comment: To find a 98% confidence interval not that we have Z = 2.3263, p = 0.38 and:

0.3304 < p < 0.4296


A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the researcher will commit a type I error if:

Incorrect


she concludes that the new diet is better when in fact the diets are equal in effectiveness.

Comment: Her hypotheses are:

Ho: New diet worse or equal to oldHa: New diet better than old

For a type I error, she needs to reject the null hypothesis when it is actually true. So the answer is:

"she concludes that the new diet is better when in fact the diets are equal in effectiveness."


In an stock portfolio selection process, a financial consultant observed the value of 40 stocks listed in the NASDAQ. The mean value of the stocks



are USD 1.22 (in millions) and the standard deviation is USD 0.17 (millions). Assuming normality, a 90% confidence interval is:

Incorrect


(1.176 , 1.264)



Your response Correct response

A sample of 600 racing cars showed that 300 cars cost over

$400,000. What is the 98% confidence interval of the true proportion of cars costing over $400,000 (3 decimals)?

(0%) < p < (0%)

A sample of 600 racing cars showed that 300 cars cost over $400,000.

What is the 98% confidence interval of the true proportion of cars costing over $400,000 (3 decimals)?

0.4525±0.01 < p < 0.5475±0.01

Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%Comment:

Our estimated mean is . Our confidence interval is .

For a 98% Confidence Interval we have z = 2.3263, so our interval is or


It was found that in a sample of 350 grandparents, 10% of them have received speeding tickets. What is the 90% confidence interval of the true

https://maple-ta.uwaterloo.ca/mapleta/web/StaRooCo000/Public_Html/Tools/NormalCalculator.htm


proportion of grandparents who have received speeding tickets? Incorrect



A clerk researched the average number of years served by 47 different judges on her court. The average number of years served was 13.11 years with a standard deviation of 7.95 years. What is the 95% confidence interval for the average number of years served by all such judges?

Incorrect


10.837 < μ < 15.383

Comment: To find a 95% CI we need the value of Z such that 97.5% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find

this value is 1.959964. The CI then is: =(10.837,

15.383)


The critical value for a left-tailed t-test for dependent samples when the degrees of freedom = 7 and α = 0.025 is : (4 decimals)

IncorrectYour

Answer:Correct

Answer:-2.365±0.001

Comment:Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0.




The Pizza Shop wanted to determine what proportion of its customers ordered only a deluxe speciality pizza. Out of 129 customers surveyed, 21 ordered a deluxe speciality pizza. What is the 99% confidence interval of the true proportion of customers who order only a deluxe speciality pizza? Incorrect



It was found that in a sample of 370 single women, 25% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of single women who have received speeding tickets?

Incorrect



A recent study of 553 Internet users in Europe found that 19% of Internet users were seniors. What is the 99% confidence interval of the true proportion of seniors in Europe who use the Internet? Incorrect


0.147< p < 0.233

Comment: For a 99% confidence interval we have Z = 2.5758 and p = 0.19 so the interval is:

0.147 < p < 0.233





A sample of 1,020 was used to estimate a proportion with 99% confidence. If p = 0.55, what was the amount of error?


0.0401

Comment:where we have used

the fact that for a 99% confidence interval, Z = 2.5758 .







The Bata Shoe Museum states that 33% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.01 of the true proportion. How large a sample is necessary?

Incorrect




IncorrectYour Answer:



Correct Answer:

Comment:



The null and alternate hypothesis of interest is:

Incorrect






from, .


A survey of 519 women shoppers found that 38% of them shop on impulse. What is the 98% confidence interval for the true proportion of

women shoppers who shop on impulse? Incorrect


0.3304 < p < 0.4296


0.3304 < p < 0.4296


A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the researcher will commit a type I error if:

Incorrect










Incorrect


(1.176 , 1.264)






(0%) < p < (0%)



0.4525±0.01 < p < 0.5475±0.01

Incorrect








It was found that in a sample of 350 grandparents, 10% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of grandparents who have received speeding tickets?

Incorrect



A clerk researched the average number of years served by 47 different judges on her court. The average number of years served was 13.11 years with a standard deviation of 7.95 years. What is the 95% confidence interval for the average number of years served by all such judges?

Incorrect


10.837 < μ < 15.383



15.383)





IncorrectYour

Answer:Correct

Answer:-2.365±0.001






It was found that in a sample of 370 single women, 25% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of single women who have received speeding tickets?

Incorrect





0.147< p < 0.233





0.147 < p < 0.233




0.0401



Question 1: Score 0/2In an stock portfolio selection process, a financial consultant observed the value of 44 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.46 (in millions) and the standard deviation is USD 0.15 (millions). Assuming normality, a 90% confidence interval is:

Incorrect


(1.423 , 1.497)






Incorrect


(395.94,404.06)









0.0227




A study on the oxygen consumption rate (OCR) of sea cucumbers involved a random sample of size 13 at 15oC and a second random sample of size 3 kept at




18oC. If one tested the hypothesis that this range of temperature had no effect on the OCR, the liberal degrees of freedom for the test statistic would be

Incorrect




Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Errorargro 2.7002

Lower 95% -14.1331

Upper 95% -2.1003


Which of the following is correct?

Incorrect





A previous analysis of celery stalks showed that the the standard deviation of their lengths is 7 millimeters. A packer wishes to find the 92% confidence interval for the average length of celery stalks. How many celery stalks must be measured to be accurate within ±7 millimeters?

Incorrect


3

Comment: The z value corresponding to this confidence level is 1.7507. We need n such that

Solving:



error of the mean. (3 decimal accuracy) IncorrectYour Answer:



SD.



A cooking school believes that 54% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.05 with 95% confidence?

Incorrect


381

Comment:Use the fact that . For a 95% confidence level Z =

1.959964, so in this case:



, solving:


We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 9 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.013. This means:

Incorrect





In a study of human mortality rate, an Actuary estimated that in US and Canada, about 68% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.01. How many claims should be included in the study?

Incorrect

Your Answer:Correct

Answer:2,176

Comment:Since we can solve for n to

get:



During the pre-flight check, Pilot Smith discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Smith decides to check the fuel level by hand, it will delay the flight by 30 minutes. If Smith decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:

i) the appropriate null hypothesis? and;ii) a type I error?

Incorrect


Null Hypothesis: assume that the warning can be ignored. Type I error: decide to check the fuel by hand when there is in fact enough fuel.

Comment:


A recent poll of 1,110 people who work indoors found that 190 of them smoke. If the researchers want to be 98% confident of their results to within 0.03, how large a sample is necessary? Incorrect









A recent poll of 1,010 people who work indoors found that 130 of them smoke. If the researchers want to be 98% confident of their results to within 0.03, how large a sample is necessary? Incorrect


Question 1: Score 0/2A recent study of 530 Internet users in Europe found that 33% of Internet users were unsupervised children. What is the 98% confidence interval of the true proportion of unsupervised children in Europe who use the Internet? Incorrect


0.2825< p < 0.3775


0.2825 < p < 0.3775


The College of Podiatrists states that 75% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 0.05 of the true proportion. How large a sample is necessary?

Incorrect




A random sample of 511 women found that 14% were

A random sample of 511 women found that 14% were going to






going to vote for a certain candidate. Find the 95% limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.) (0%) < p < (0%)

vote for a certain candidate. Find the 95% limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.)0.1099±0.01 < p < 0.1701±0.01

Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%Comment:To find a 95% confidence interval not that we have Z = 1.96, p = 0.14 and:

0.1099 < p < 0.1701




0.0204




An agronomist measured the height of 110 Canola plants. The mean height was 220 cm and the standard deviation was 16 cm. Calculate the standard error of the mean. (3 decimal accuracy) IncorrectYour Answer:




SD.




A Type I (false positive) error would occur if:

Incorrect

Your Answer:

Correct Answer:

We conclude that larger nests had larger eggs (on average) when in fact there is no difference in the mean.

Comment: Type I error occurs when a true null hypothesis is rejected in favour of the alternate hypothesis. In this case, Type I error means that we reject the hypothesis that larger nests have the same size of eggs as smaller nests.



True Status

Innocent Guilty

Examiner'sDecision

Innocent 15 24

Guilty 135 126


Incorrect


0.9

Comment:




Incorrect

Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Upper 95% -2.1003







Incorrect



A cooking school believes that 47% of applicants to that school have parents who were alumni. How large a sample is needed to estimate the



true proportion of students who have parents who were alumni to within 0.03 with 95% confidence?

Incorrect


1,063



, solving:



True Status

Innocent Guilty

Examiner'sDecision

Innocent 10 30

Guilty 90 70

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we could estimate the probability of making a type II error as:

Incorrect


0.3

Comment: Type II error is when a false null hypothesis is not rejected, also known as a "false negative". Given the null hypothesis that the suspect is innocent, we wish to test the probability that the Examiner judges a guilty suspect to be innocent. Based on the given numbers, this is

.







A report states that 42% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks

to within 0.03 with 90% confidence? Incorrect


732

Comment: For a 90% confidence level we have Z = 1.6449. Use the fact that

by substituting in the known values and solving for n :

,


An agronomist measured the height of 117 Soybean plants. The mean height was 233 cm and the standard deviation was 12 cm. Calculate the

standard error of the mean. IncorrectYour Answer:Correct Answer: 1.1094Comment:


SD.




A quality control engineer wants to estimate the proportion of defective parts that are being manufactured by his company to within 2.5%. A sample of 550 components showed that 15 were defective. How large a sample is needed to estimate the true proportion of defective parts with 99% confidence?

Incorrect

Your Answer:Correct Answer: 282

Comment:Use the fact that

Question 1: Score 0/2A medical researcher is interested in whether patients' left arms or right arms are longer. If 13 patients participate in this study (so that 13 left arms and 13 right arms are measured), how many degrees of freedom should the researcher use in her t-test critical value?

Incorrect


12




Incorrect

Your Answer:



Correct Answer:

460


get:



True Status

Innocent Guilty

Examiner'sDecision

Innocent 20 28

Guilty 130 122


Incorrect


0.867

Comment:




Incorrect

Your Answer:


Correct Answer: 553Comment:

Use the equation: . For a 95% confidence interval we have c = 1.96,

so:





During the pre-flight check, Pilot Jones discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Jones decides to check the fuel level by hand, it will delay the flight by 50 minutes. If Jones decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:


Incorrect



Comment:




Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Upper 95% -2.1003



Incorrect





In a study of human mortality rate, an Actuary estimated that in US and Canada, about 92% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.012. How many claims should

Incorrect


be included in the study?Your Answer:

Correct Answer:

512


get:


The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 109 customers surveyed, 40 ordered

cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Incorrect






A previous analysis of paper boxes showed that the the standard deviation of their lengths is 7 millimeters. A pallet manufacturer wishes to find the 96% confidence interval for the average length of paper boxes. How many paper boxes must be measured to be accurate within ±4 millimeters?

Incorrect


13





Solving:


A previous analysis of celery stalks showed that the the standard deviation of their lengths is 10 millimeters. A bag manufacturer wishes to find the 92% confidence interval for the average length of celery stalks. How many celery stalks must be measured to be accurate within ±3 millimeters?

Incorrect


34


Solving:


A random sample of 506 seniors found that 23% were going to vote for a certain candidate. Find the 95% limit for the population

proportion of seniors who will vote for that candidate. IncorrectYour Answer:Correct Answer: 0.1933 < p < 0.2667Comment: To find a 95% confidence interval not that we have Z = 1.96, p = 0.23 and:

0.1933 < p < 0.2667


An agronomist measured the height of 110 Soybean plants. The mean height was 242 cm and the standard deviation was 18 cm. Calculate the standard error of the mean. Incorrect

Your Answer:



Correct Answer: 1.7162Comment:


SD.




True Status

Innocent Guilty

Examiner'sDecision

Innocent 13 23

Guilty 87 77


Incorrect


0.23


.

Question 1: Score 0/2A clerk researched the average number of years served by 50 different judges on his court. The average number of years served

was 13.98 years with a standard deviation of 7.77 years. What is the 96% confidence interval for the average number of years served by all such judges?

Incorrect


11.723 < μ < 16.237

Comment: To find a 96% CI we need the value of Z such that 98% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find



16.237)



A sample of 360 racing cars showed that 86 cars cost over $300,000. What is the 98% confidence interval of the true proportion of cars costing over $300,000 (3 decimals)?

(0%) < p < (0%)

A sample of 360 racing cars showed that 86 cars cost over $300,000. What is the 98% confidence interval of the true proportion of cars costing over $300,000 (3 decimals)?

0.1866±0.01 < p < 0.2912±0.01

Incorrect






Incorrect





Comment:Use the equation: . For a 95% confidence interval we have c = 1.96,

so:


A shift manager wants to estimate the proportion of defective parts that are being manufactured by her company to within 2.5%. A sample of 500 components showed that 16 were defective. How large a sample is needed to estimate the true proportion of defective parts with 99% confidence?

Incorrect


Comment:Use the fact that


In a study of stock options, a sample of 198 stock options were observed and 54 were discovered to have a final negative payoff. Construct a 98% confidence interval for the relative frequency of those stock options with negative payoff.

Incorrect


(0.1993 , 0.3467)

Comment:First . Then for a 98% confidence interval we have Z =

2.3263 so the interval is:





Incorrect

Your Answer:Correct

Answer:1,395


get:



IncorrectYour

Answer:Correct

Answer:-2.365±0.001



A report states that 84% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.01 with 98% confidence? Incorrect

Your Answer:



Correct Answer:

7,273



,



Incorrect






from, .



Incorrect


(376.81,383.19)




A cooking school believes that 77% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.01 with 95% confidence? (4 decimal accuracy)

Incorrect

Your Answer:Correct

Answer:6,803±1



, solving:

You need to round off to the integer since you cannot take a fraction of a student.



Incorrect

Your



Answer:Correct Answer:









A study on the oxygen consumption rate (OCR) of sea cucumbers involved a random sample of size 13 at 15oC and a second random sample of size 6 kept at 18oC. If one tested the hypothesis that this range of temperature had no effect on the OCR, the liberal degrees of freedom for the test statistic would be

Incorrect



In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used? Incorrect

Your Answer:Correct Answer: 99%Comment:


Question 1: Score 0/2A study on the oxygen consumption rate (OCR) of sea cucumbers involved a random sample of size 13 at 15oC and a second random sample of size 10 kept at 18oC. If one tested the hypothesis that this range of temperature had no effect on the OCR, the liberal degrees of freedom for the test statistic would be

Incorrect



A trade school believes that 25% of applicants to that school have parents who were alumni. How large a sample is needed to estimate the true proportion of students who have parents who were alumni to within 0.03 with 95% confidence?

Incorrect


800



, solving:



Incorrect



Comment: The two-sample t-test tests the null hypothesis that the mean of two populations is the same. In this case the two populations are the animals while eating the old feed, and the animals while eating the new feed. The p-value is the probability of a more rare (more difference in weight gained) occurrence, given that the null hypothesis is


true.


We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 6.7 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.084. This means:

Incorrect









Incorrect






Incorrect


(417.11,422.89)




A survey of 565 men shoppers found that 41% of them shop on impulse. What is the 99% confidence interval for the true proportion of men

shoppers who shop on impulse? Incorrect


0.3567 < p < 0.4633


0.3567 < p < 0.4633



Your Answer:




Correct Answer: 0.066 < p < 0.133Comment:


A study of elephants wishes to determine the average weight of a certain subspecies of elephants. The standard deviation of the population is 2,074 kilograms. How many elephants need to be weighed so we can be 95% confident to be accurate within 308 kilograms?

Incorrect


174

Comment: For a 95% CI we need to find z such that 97.5% of the standard normal curve area

lies to the left of z. This value is z = 1.96. Since the CI is we need to

find n such that . Solving:



True Status

Innocent Guilty

Examiner'sDecision

Innocent 18 27

Guilty 132 123


Incorrect


0.88

Comment:




During the pre-flight check, Pilot Singh discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Singh decides to check the fuel level by hand, it will delay the flight by 55 minutes. If Singh decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:


Incorrect



Comment:



Incorrect


(476.74,483.26)









Researchers tested patients fitted with an automated insulin pump to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 48% of these tests.

Construct a 95% Conficence Interval (4 decimal accuracy). Hint: use the General confidence interval for p.

( (0%) , (0%) )

Researchers tested patients fitted with an automated insulin pump to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 48% of these tests.


( 0.4394±0.001 , 0.5206±0.001 )

Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%Comment:The confidence interval would be:

=

= (0.4394, 0.5206)

Question 1: Score 0/2A study of 93 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 11 m. Which of the following is the 99% confidence interval for the mean length per bolt of carpet? Incorrect


(179.06, 184.94)







A researcher is going to conduct an experiment in order to compare two treatments – a new treatment and an old treatment. The researcher would like to see whether there is sufficient evidence to say that the new treatment is better than the old treatment. In this problem, the researcher will commit a type I error if:

Incorrect


she concludes that the new treatment is better when in fact the treatments are equal in effectiveness.


Ho: New treatment worse or equal to oldHa: New treatment better than old


"she concludes that the new treatment is better when in fact the treatments are equal in effectiveness."


A college believes that 60% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.05 with 95% confidence? (4 decimal accuracy)

Incorrect

Your Answer:Correct

Answer:369±1



, solving:









The Pizza Shop wanted to determine what proportion of its customers ordered only vegetarian pizza. Out of 127 customers surveyed, 23 ordered

vegetarian pizza. What is the 99% confidence interval of the true proportion of customers who order only vegetarian pizza? Incorrect




Incorrect








Incorrect


(306.07,313.93)





Incorrect


(1.134 , 1.206)






During the pre-flight check, Pilot Jones discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Jones

decides to check the fuel level by hand, it will delay the flight by 50 minutes. If Jones decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:


Incorrect



Comment:



Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Upper 95% -2.1003

Incorrect






A clerk researched the average number of years served by 49 different judges on the District Court. The average number of years

served was 13.26 years with a standard deviation of 7.95 years. What is the 96% confidence interval for the average number of years served by all such judges?

Incorrect


10.928 < μ < 15.592

Comment: To find a 96% CI we need the value of Z such that 98% of the normal curve area lies to the left of it. Use a standard normal table or the calculator provided to find


15.592)


A cooking school believes that 17% of applicants to that school have parents who cannot help the student financially. How large a sample is needed to estimate the true proportion of students who have parents who cannot help the student financially to within 0.02 with 95% confidence?

Incorrect


1,355





, solving:



Incorrect


(0.4254 , 0.6466)

Comment:First . Then for a 98% confidence interval we have Z =

2.3263 so the interval is:


An agronomist measured the height of 109 Corn plants. The mean height was 241 cm and the standard deviation was 14 cm. Calculate the standard error of the mean. (3 decimal accuracy) IncorrectYour Answer:




Question 1: Score 0/2A college believes that 71% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.05 with 95% confidence?

Incorrect

Your Answer:Correct 316



Answer:Comment:

Use the fact that . For a 95% confidence level Z =


, solving:



Incorrect






from, .


A college believes that 79% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 0.04 with 95% confidence? (4 decimal accuracy)

Incorrect

Your Answer:Correct

Answer:398±1



, solving:




Incorrect

Your Answer:Correct

Answer:241


get:



Your Answer:




Correct Answer:

0.4026< p < 0.4974


0.4026 < p < 0.4974


The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 116 customers surveyed, 21 ordered

cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Incorrect





(179.9, 186.1)






Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Upper 95% -2.1003



Incorrect





In a study of human mortality rate, an Actuary estimated that in US and Canada, about 76% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be 0.012. How many claims should

Incorrect


be included in the study?Your Answer:

Correct Answer:

1,267


get:





SD.




Incorrect



A cooking school believes that 35% of applicants to that school have parents who cannot help the student financially. How large a sample is needed to estimate the true proportion of students who have parents who cannot help the student financially to within 0.01 with 95% confidence? (4 decimal accuracy)

Incorrect

Your Answer:


Correct Answer:

8,739±1



, solving:




Incorrect





A study of rhinos wishes to determine the average weight of a certain family group of rhinos. The standard deviation of the population is 2,086 kilograms. How many rhinos need to be weighed so we can be 92% confident to be accurate within 340 kilograms?

Incorrect


115

Comment: For a 92% CI we need to find z such that 96% of the standard normal curve area


lies to the left of z. This value is z = 1.7507. Since the CI is we need to

find n such that . Solving:



True Status

Innocent Guilty

Examiner'sDecision

Innocent 11 29

Guilty 89 71


Incorrect


0.29


.

Question 1: Score 0/2In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used? Incorrect




During the pre-flight check, Pilot VanDerBoek discovers a minor problem - a warning light indicates that the fuel guage may be broken. If VanDerBoek decides to check the fuel level by hand, it will delay the flight by 30 minutes. If VanDerBoek decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:


Incorrect



Comment:



IncorrectYour

Answer:Correct

Answer:-2.365±0.001




True Status

Innocent Guilty

Examiner'sDecision

Innocent 15 20

Guilty 135 130

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we

Incorrect

could estimate the probability of making a type I error as:Your Answer:Correct Answer:

0.9

Comment:




Incorrect


(0.3805 , 0.5655)

Comment:First . Then for a 95% confidence interval we have Z = 1.96

so the interval is:


An agronomist measured the height of 136 Corn plants. The mean height was 210 cm and the standard deviation was 11 cm. Calculate the standard error of the mean. Incorrect

Your Answer:Correct Answer: 0.9432Comment:


SD.




A report states that 82% of home owners had a gazebo. How large a sample is needed to estimate the true proportion of home owners who

have gazebos to within 0.02 with 98% confidence? Incorrect


1,997



,




(176.3, 183.7)



An agronomist measured the height of 149 Canola plants. The mean height was 221 cm and the standard deviation was 19 cm. Calculate the standard error of the mean. Incorrect

Your Answer:Correct Answer: 1.5565



Comment:The Standard Error is , where s is sample

SD.




Incorrect

Your Answer:Correct

Answer:2,356


get:





A random sample of 530 seniors found that 22% were going to vote for a certain candidate. Find the 95% limit for the population

proportion of seniors who will vote for that candidate. IncorrectYour Answer:Correct Answer: 0.1847 < p < 0.2553Comment: To find a 95% confidence interval not that we have Z = 1.96, p = 0.22 and:




0.1847 < p < 0.2553


A recent study of 538 Internet users in Europe found that 40% of Internet users were unsupervised children. What is the 99% confidence interval of the true proportion of unsupervised children in Europe who use the Internet? Incorrect


0.3456< p < 0.4544


0.3456 < p < 0.4544



True Status

Innocent Guilty

Examiner'sDecision

Innocent 15 29

Guilty 85 71

If the hypotheses were H: suspect is innocent vs A: suspect is guilty, then we

Incorrect


could estimate the probability of making a type II error as:Your Answer:Correct Answer:

0.29


.



Incorrect






Incorrect


(335.7,344.3)







0.0245




During the pre-flight check, Pilot Smith discovers a minor problem - a warning light indicates that the fuel guage may be broken. If Smith decides to check the fuel level by hand, it will delay the flight by 30 minutes. If Smith decides to ignore the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:


Incorrect



Comment:



The Pizza Shop wanted to determine what proportion of its customers ordered only a deluxe speciality pizza. Out of 105 customers surveyed, 38

ordered a deluxe speciality pizza. What is the 99% confidence interval of the true proportion of customers who order only a deluxe speciality pizza? Incorrect







Incorrect



A Type I (false positive) error would occur if:Your Answer:Correct Answer:

We conclude that larger nests had larger eggs (on average) when in fact there is no difference in the mean.

Comment: Type I error occurs when a true null hypothesis is rejected in favour of the alternate hypothesis. In this case, Type I error means that we reject the hypothesis that larger nests have the same size of eggs as smaller nests.


A medical researcher is interested in whether patients' left arms or right arms are longer. If 13 patients participate in this study (so that 13 left arms and 13 right arms are measured), how many degrees of freedom should the researcher Incorrect

use in her t-test critical value?Your Answer:Correct Answer:

12






(0%) < p < (0%)



0.4624±0.01 < p < 0.5558±0.01

Incorrect






Incorrect








from, .






True Status

Innocent Guilty

Examiner'sDecision

Innocent 18 25

Guilty 82 75


Incorrect


0.25


.




(177.97, 186.03)








SD.




Incorrect


(0.2133 , 0.3447)

Comment:First . Then for a 95% confidence interval we have Z = 1.96

so the interval is:


A public defender researched the average number of years served by 46 different judges on the District Court. The average number of years served was 13.85 years with a standard deviation of 7.33 years. What is the 95% confidence interval for the average number of years served by all such judges?

Incorrect




11.732 < μ < 15.968



15.968)

Question 1: Score 0/2The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 119 customers surveyed, 39 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Incorrect




Researchers tested patients fitted with a cardiac pacemaker to see if use of a cellular telephone interferes with the operation of the device. There were 518 tests conducted for one type of cellular telephone; interference with the device was found in 49% of these tests.


( (0%) , (0%) )



( 0.4389±0.001 , 0.5411±0.001 )

Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%Comment:The confidence interval would be:




=

= (0.4389, 0.5411)


A researcher is going to conduct an experiment in order to compare two treatments – a new treatment and an old treatment. The researcher would

like to see whether there is sufficient evidence to say that the new treatment is better than the old treatment. In this problem, the researcher will commit a type I error if:

Incorrect


she concludes that the new treatment is better when in fact the treatments are equal in effectiveness.


Ho: New treatment worse or equal to oldHa: New treatment better than old


"she concludes that the new treatment is better when in fact the treatments are equal in effectiveness."



A random sample of 544 men found that 35% were going to vote for a certain candidate. Find the 98% limit for the population proportion of men who will vote for that candidate. (3 decimal accuracy.) (0%) < p < (0%)

A random sample of 544 men found that 35% were going to vote for a certain candidate. Find the 98% limit for the population proportion of men who will vote for that candidate. (3 decimal accuracy.)0.3024±0.01 < p < 0.3976±0.01

Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%Comment:To find a 98% confidence interval not that we have Z = 2.3263, p = 0.35 and:



0.3024 < p < 0.3976






Incorrect




Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Incorrect


Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Upper 95% -2.1003







True Status

Innocent Guilty

Examiner'sDecision

Innocent 19 22

Guilty 81 78


Incorrect


0.22


.


Researchers tested patients fitted with an automated insulin pump to see if use of a cellular telephone interferes with the operation of the device. There were 600 tests conducted for one type of cellular telephone; interference with the device was found in 31% of these tests. Which of the following is a 98% Confidence Interval? Hint: use the General confidence interval for p.

Incorrect


=

= (0.2661, 0.3539)



Incorrect







from, .




(177.71, 182.29)









An agronomist measured the height of 146 Soybean plants. The mean height was 247 cm and the standard deviation was 19 cm. Calculate the standard error of the mean. Incorrect

Your Answer:Correct Answer: 1.5725




Comment:The Standard Error is , where s is sample

SD.



A cooking school believes that 28% of applicants to that school have parents who cannot help the student financially. How large a sample is needed to estimate the true proportion of students who have parents who cannot help the student financially to within 0.01 with 95% confidence?

Incorrect


Use the fact that . For a 95% confidence level Z =


, solving:

STAT 202 1121 Introductory Statistics for Scientists : Quiz 10

Welcome Adam Joseph Amador

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STAT 202 1121 Introductory Statistics for Scientists, Quiz 10Adam Joseph Amador, 3/27/12 at 11:37 AM


Question 1: Score 0/2A study of 95 bolts of carpet showed that their average length was 185 meters. The standard deviation of the population is 10 m. Which of the following is the 97% confidence interval for the mean length per bolt of carpet?


(182.77, 187.23)




Incorrect


Comment:




5,912

Comment:

Use the fact that . For a 95% confidence level Z = 1.959964, so in this

case:

, solving:



Incorrect






Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean egg

size in smaller nests, or . The null hypothesis is that the egg size is the same no matter

which nest the egg came from from, .


12 squirrels were found to have an average weight of 490 grams with a sample standard deviation of 6. Find the 95% confidence interval of the true mean weight (assume the t-student distribution).

Incorrect

Your Answer:


Correct Answer:

(486.19,493.81)

Comment:

Use the formula . With degrees of freedom = 11 we have t0.025 = 2.201 so our

confidence interval is



Which of the following is a 99% Confidence Interval? Hint: use the General confidence interval for p.

Incorrect


=

= (0.3364, 0.4436)


In an stock portfolio selection process, a financial consultant observed the value of 43 stocks listed in the NASDAQ. The mean value of the stocks are USD 1.6 (in millions) and the standard deviation is USD 0.2 (millions). Assuming normality, a 90% confidence interval is: Incorrect

Your Answer:Correct Answer: (1.55 , 1.65)Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:


A retailer wants to estimate with 99% confidence the number of people who buy at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the true proportion. The minimum sample size necessary would be 1,100. Incorrect

Your Answer:Correct Answer: FalseComment:



Incorrect




Your Answer:Correct

Answer:-2.365±0.001

Comment:Looking at a t-table, with the specified degrees of freedom and , the value 2.365 shows up. Since

the test is a one-sided left-tailed test, this limiting value should be negative because the tail is to the left of 0.


A study on the oxygen consumption rate (OCR) of sea cucumbers involved a random sample of size 12 at 15oC and a second random sample of size 3 kept at 18oC. If one tested the hypothesis that this range of temperature had no effect on the OCR, the liberal degrees of freedom for the test statistic would be Incorrect





(1.683 , 1.777)



A report states that 18% of home owners had a gazebo. How large a sample is needed to estimate the true proportion of home owners who have gazebos to within 0.04 with 98% confidence?

Incorrect


499

Comment:

For a 98% confidence level we have Z = 2.3263. Use the fact that by

substituting in the known values and solving for n :

,


A recent poll of 1,190 people who work indoors found that 200 of them smoke. If the researchers want to be 98% confident of their results to within 0.04, how large a sample is necessary?

IncorrectYour Answer:Correct Answer: 473




Comment:


A recent study of 518 Internet users in Europe found that 20% of Internet users were unsupervised children. What is the 99% confidence interval of the true proportion of unsupervised children in Europe who use the Internet?

IncorrectYour Answer:Correct Answer: 0.1547< p < 0.2453Comment:

For a 99% confidence interval we have Z = 2.5758 and p = 0.2 so the interval is:

0.1547 < p < 0.2453


A medical researcher is interested in whether patients' left arms or right arms are longer. If 11 patients participate in this study (so that 11 left arms and 11 right arms are measured), how many degrees of freedom should the researcher use in her t-test critical value? Incorrect


10


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(182.77, 187.23)




Incorrect


Comment:




5,912

Comment:


case:

, solving:






Incorrect



Comment: The researcher is interested in determining whether larger birds lay larger eggs. This statement describes the alternate hypothesis that the mean egg size in larger nests is larger than the mean

egg size in smaller nests, or . The null hypothesis is that the egg size is the same no

matter which nest the egg came from from, .


12 squirrels were found to have an average weight of 490 grams with a


sample standard deviation of 6. Find the 95% confidence interval of the true mean weight (assume the t-student distribution).

Incorrect


(486.19,493.81)

Comment:






Incorrect


=

= (0.3364, 0.4436)



Your Answer:Correct Answer: (1.55 , 1.65)Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:









IncorrectYour

Answer:Correct

Answer:-2.365±0.001









(1.683 , 1.777)




Incorrect


499

Comment:



,





IncorrectYour Answer:Correct Answer: 473Comment:





0.1547 < p < 0.2453




10


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(182.77, 187.23)




Incorrect


Comment:




5,912

Comment:


case:

, solving:






Incorrect








Incorrect


(486.19,493.81)

Comment:

Use the formula . With degrees of freedom = 11 we have t0.025 = 2.201 so

our confidence interval is




Incorrect


=

= (0.3364, 0.4436)



IncorrectYour Answer:Correct Answer: (1.55 , 1.65)Comment: Here we use the provided Normal calculator to get F(0.95) = 1.6449, so the 90% CI is:










IncorrectYour

Answer:Correct

Answer:-2.365±0.001









(1.683 , 1.777)




Incorrect


499

Comment:



,










0.1547 < p < 0.2453




10


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(182.77, 187.23)




Incorrect


Comment:




5,912

Comment:


case:

, solving:






Incorrect








Incorrect


(486.19,493.81)

Comment:






Incorrect


=

= (0.3364, 0.4436)













IncorrectYour

Answer:Correct

Answer:-2.365±0.001









(1.683 , 1.777)




Incorrect


499

Comment:



,










0.1547 < p < 0.2453




10


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(182.77, 187.23)




Incorrect


Comment:




5,912

Comment:


case:

, solving:






Incorrect








Incorrect


(486.19,493.81)

Comment:






Incorrect


=

= (0.3364, 0.4436)













IncorrectYour

Answer:Correct

Answer:-2.365±0.001









(1.683 , 1.777)




Incorrect


499

Comment:



,










0.1547 < p < 0.2453




10


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(182.77, 187.23)




Incorrect


Comment:




5,912

Comment:


case:

, solving:






Incorrect








Incorrect


(486.19,493.81)

Comment:






Incorrect


=

= (0.3364, 0.4436)













IncorrectYour

Answer:Correct

Answer:-2.365±0.001









(1.683 , 1.777)




Incorrect


499

Comment:



,










0.1547 < p < 0.2453




10


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Question 1: Score 0/2A previous analysis of celery stalks showed that the the standard deviation of their lengths is 8 millimeters. A pallet manufacturer wishes to find the 91% confidence interval for the average length of celery stalks. How many celery stalks must be measured to be accurate within ±7 millimeters? Incorrect


4

Comment:

The z value corresponding to this confidence level is 1.6954. We need n such that

Solving:






IncorrectYour

Answer:Correct

Answer:-2.365±0.001




An agronomist measured the height of 139 Corn plants. The mean height was 228 cm and the standard deviation was 19 cm. Calculate the standard error of the mean.

IncorrectYour Answer:Correct Answer: 1.6116



Comment:





A random sample of 514 women found that 10% were going to vote for a certain candidate. Find the

99% limit for the population proportion of women who will vote for that candidate. (3 decimal accuracy.)

(0%) < p < (0%)

A random sample of 514 women found that 10% were going to vote for a certain candidate. Find the 99% limit for

the population proportion of women who will vote for that candidate. (3 decimal accuracy.)

0.0659±0.01 < p < 0.1341±0.01 Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%

Comment:

To find a 99% confidence interval not that we have Z = 2.5758, p = 0.1 and:

0.0659 < p < 0.1341


A survey of 535 men shoppers found that 42% of them shop on impulse. What is the 98% confidence interval for the true proportion of men shoppers who shop on impulse?

IncorrectYour Answer:Correct Answer: 0.3704 < p < 0.4696Comment: To find a 98% confidence interval not that we have Z = 2.3263, p = 0.42 and:

0.3704 < p < 0.4696






Incorrect


(285.31,294.69)

Comment:





A random sample of 512 seniors found that 16% were going to vote for a certain candidate. Find the 95% limit for the population proportion of seniors who will vote for that candidate. (3 decimal accuracy.)

(0%) < p < (0%)

A random sample of 512 seniors found that 16% were going to vote for a certain candidate. Find the 95% limit for the population proportion of seniors who will vote for that candidate. (3 decimal accuracy.)

0.1282±0.01 < p < 0.1918±0.01 Incorrect

Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%

Comment:

To find a 95% confidence interval not that we have Z = 1.96, p = 0.16 and:

0.1282 < p < 0.1918


A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the researcher will commit a type I error if: Incorrect

Your Answer:Correct Answer: she concludes that the new diet is better when in fact the diets are equal in effectiveness.Comment: Her hypotheses are:









Incorrect


Comment:







13





(0.2065 , 0.3815)

Comment:

First . Then for a 99% confidence interval we have Z = 2.5758 so the

interval is:




(0.3744 , 0.6096)




Comment:


interval is:


We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.089. This means: Incorrect




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Question 1: Score 0/2We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 15 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.027. This means: Incorrect






IncorrectYour

Answer:Correct

Answer:-2.365±0.001




A random sample of 594 printers discovered that 53 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses.

IncorrectYour Answer:Correct Answer: 0.059 < p < 0.119Comment:


Researchers tested patients fitted with a remote blood glucose monitor to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 14% of these tests.


Incorrect


=

= (0.1029, 0.1771)



Incorrect





(0.2493 , 0.3567)

Comment:


interval is:


Suppose a study is being planned to estimate the relative frequency of companies that are labeled as risky. What sample size is needed so that the standard error will be no larger than 0.018.

Hint: Find p that maximizes the standard error. Incorrect


Since , SE is maximized by setting . Do so and solve for n:


An agronomist measured the height of 108 Canola plants. The mean height was 245 cm and the standard deviation was 20 cm. Calculate the standard error of the mean.

IncorrectYour Answer:Correct Answer: 1.9245Comment:











13




Incorrect


(483.86,496.14)

Comment:





True Status

Innocent Guilty

Examiner'sDecision

Innocent 13 28

Guilty 137 122


Incorrect


0.913

Comment: Type I error is when a true null hypothesis is rejected. In this case, this is the probability of the Examiner judging an innocent suspect to be guilty. The probability is:



A survey of 516 depressed shoppers found that 36% of them shop on impulse. What is the 99% confidence interval for the true proportion of depressed shoppers who shop on impulse?

IncorrectYour Answer:Correct Answer: 0.3056 < p < 0.4144


0.3056 < p < 0.4144


A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the researcher will commit a type I error if: Incorrect








A report states that 83% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.03 with 95% confidence?

Incorrect


602

Comment:



,




A study of 94 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 13 m. Which of the following is the 99% confidence interval for the mean length per bolt of carpet?


(178.55, 185.45)


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Question 1: Score 0/2We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 15 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.027. This means: Incorrect







IncorrectYour

Answer:Correct

Answer:-2.365±0.001




A random sample of 594 printers discovered that 53 of them were being used in small businesses . Find the 99% limit for the population proportion of printers that are used in small businesses.

IncorrectYour Answer:Correct Answer: 0.059 < p < 0.119Comment:


Researchers tested patients fitted with a remote blood glucose monitor to see if use of a cellular telephone interferes with the operation of the device. There were 581 tests conducted for one type of cellular telephone; interference with the device was found in 14% of these tests.


Incorrect


=

= (0.1029, 0.1771)




(0.2493 , 0.3567)

Comment:


interval is:










An agronomist measured the height of 108 Canola plants. The mean height was 245 cm and the standard deviation was 20 cm. Calculate the standard error of the mean.

IncorrectYour Answer:Correct Answer: 1.9245Comment:










Your Answer:Correct 13

Answer:Comment: The attribute under study is the difference in arm length from a person's left to right arm. Since there

is one observation of this per individual in the population, the degrees of freedom is 14 - 1 = 13



Incorrect


(483.86,496.14)

Comment:





True Status

Innocent Guilty

Examiner'sDecision

Innocent 13 28

Guilty 137 122


Incorrect


0.913

Comment: Type I error is when a true null hypothesis is rejected. In this case, this is the probability of the Examiner judging an innocent suspect to be guilty. The probability is:


A survey of 516 depressed shoppers found that 36% of them shop on impulse. What is the 99% confidence interval for the true proportion of depressed shoppers who shop on impulse?

IncorrectYour Answer:Correct Answer: 0.3056 < p < 0.4144Comment: To find a 99% confidence interval not that we have Z = 2.5758, p = 0.36 and:



0.3056 < p < 0.4144


A researcher is going to conduct an experiment in order to compare two diets – a new diet and an old diet. The researcher would like to see whether there is sufficient evidence to say that the new diet is better than the old diet. In this problem, the

researcher will commit a type I error if: Incorrect

Your Answer:Correct Answer: she concludes that the new diet is better when in fact the diets are equal in effectiveness.Comment: Her hypotheses are:





A report states that 83% of home owners had a deck. How large a sample is needed to estimate the true proportion of home owners who have decks to within 0.03 with 95% confidence?

Incorrect


602

Comment:



,



A study of 94 bolts of carpet showed that their average length was 182 meters. The standard deviation of the population is 13 m. Which of the following is the 99% confidence interval for the mean length per bolt of carpet?


(178.55, 185.45)


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(181.86, 188.14)

Comment: For a 98% CI we need to find the z value for the standard normal for which 99% of the graph area lies to the left. This is 2.3263 . The CI is:




A pharmacist is planning to estimate the mean level of a certain drug in a lab. The

pharmacist wanted the estimate to be within 6 mg/dLi or less with 95% confidence.

The pharmacist also believes that the standard deviation of the drug level is probably about 43 mg/dLi. How large a sample should the pharmacist need to take? Incorrect


Use the equation: . For a 95% confidence interval we have c = 1.96, so:


A previous analysis of cam shafts showed that the the standard deviation of their lengths is 9 millimeters. A bag manufacturer wishes to find the 98% confidence interval for the average length of cam shafts. How many cam shafts must be measured to be accurate within ±5 millimeters? Incorrect


18

Comment:


Solving:


The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 134 customers surveyed, 21 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? Incorrect



A researcher is going to conduct an experiment in order to compare two drugs – a new drug and an old drug. The researcher would like to see whether there is sufficient evidence to say that the new drug is better than the old drug. In this problem, the researcher will commit a type I error if: Incorrect

Your Answer:Correct Answer: she concludes that the new drug is better when in fact the drugs are equal in effectiveness.Comment: Her hypotheses are:

Ho: New drug worse or equal to oldHa: New drug better than old


"she concludes that the new drug is better when in fact the drugs are equal in effectiveness."












Incorrect


=

= (0.3282, 0.4318)


In a sample of 640 mice, a biologist found that 64% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less.

IncorrectYour Answer:Correct Answer: 0.603% < p < 0.677%Comment:



IncorrectYour

Answer:Correct

Answer:-2.365±0.001






An agronomist measured the height of 144 Wheat plants. The mean height was 209 cm and the standard deviation was 14 cm. Calculate the standard error of the mean. (3 decimal accuracy)

IncorrectYour Answer:





It was found that in a sample of 390 teenage boys, 90% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who have received speeding tickets?

Incorrect





There was only a 4% chance of observing an increase greater than 5 kg (assuming the null hypothesis was true).




Incorrect


Comment:


In a study of human mortality rate, an Actuary estimated that in US and Canada, about 65% (fictional figures) of life insurance claims resulted from accidental deaths. Suppose a study is being planned to estimate the relative frequency of claims in Canada, and it is desired that the standard error of the estimated relative frequency should be Incorrect



0.018. How many claims should be included in the study?Your Answer:

Correct Answer:

703

Comment:

Since we can solve for n to

get:


In a sample of 855 bartenders, 48% heard complaints from patrons about work. If the margin of error was 4.4%, what was the confidence level that was used?

IncorrectYour Answer:Correct Answer: 99%Comment:

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Question 1: Score 2/2In a sample of 640 mice, a biologist found that 79% were able to run a maze in 30 seconds or less. Find the 95% limit for the population proportion of mice who can run that maze in 30 seconds or less.

CorrectYour Answer: 0.757% < p < 0.82%Comment:





IncorrectYour Answer: 1,483Correct Answer: 1,349Comment:


A previous analysis of cam shafts showed that the the standard deviation of their lengths is 8 millimeters. A bag manufacturer wishes to find the 97% confidence interval for the average length of cam shafts. How many cam shafts must be measured to be accurate within ±7 millimeters? Correct

Your Answer:

6

Comment:


Solving:


An agronomist measured the height of 120 Wheat plants. The mean height was 228 cm and the standard deviation was 20 cm. Calculate the standard error of the mean.

CorrectYour Answer: 1.8257Comment:





Rat: 1 2 3 4 5 6

Poisoned Group: 12.2 16.9 25.0 22.4 8.5 20.6

Control Group: 11.1 12.1 9.3 6.6 9.6 8.2



Estimate -8.1167 -3.006 10 0.0132

Std Error 2.7002

Lower 95% -14.1331

Correct



Upper 95% -2.1003



Your Answer:




An agronomist measured the height of 121 Corn plants. The mean height was 210 cm and the standard deviation was 14 cm. Calculate the standard error of the mean.

CorrectYour Answer: 1.2727Comment:




We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.006. This means: Correct

Your Answer:




We wish to test if a new feed increases the mean weight gain compared to an old feed. At the conclusion of the experiment it was found that the new feed gave a 8.4 kg bigger gain than the old feed. A two-sample t-test with the proper one-sided alternative was done and the resulting p-value was 0.067. This means: Correct

Your Answer:




During the pre-flight check, Pilot VanDerBoek discovers a minor problem - a warning light indicates that the fuel guage may be broken. If VanDerBoek decides to check the fuel level by hand, it will delay the flight by 40 minutes. If VanDerBoek decides to ignore

the warning, the aircraft may run out of fuel before it gets to Gimli. In this situation, what would be:

Correct


Your Answer:


Comment:


Your response

Researchers tested patients fitted with a medical telemetry unit to see if use of a cellular telephone interferes with the operation of the device. There were 535 tests conducted for one type of cellular telephone; interference with the device was found in 37% of these tests.


( 0.316 (50%) , 0.424 (50%) )

Correct

Comment:

The confidence interval would be:

=

= (0.3162, 0.4238)


A medical researcher is interested in whether patients' left arms or right arms are longer. If 13 patients participate in this study (so that 13 left arms and 13 right arms are measured), how many degrees of freedom should the researcher use in her t-test critical value? Correct

Your Answer:

12




Correct

Your Answer:

(344.8,355.2)

Comment:







Your Answer: (1.037, 1.099)Correct Answer:

(1.021 , 1.099)




IncorrectYour Answer: 628Correct Answer: 215Comment:


An agronomist measured the height of 143 Wheat plants. The mean height was 218 cm and the standard deviation was 10 cm. Calculate the standard error of the mean. (3 decimal accuracy)

CorrectYour Answer: 0.836

Comment:





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