Inference and Confidence Intervals

Introduction to Inference

Moore IPS Chapter 6

© 2012 W.H. Freeman and Company

A researcher wants to know if the average time in jail for robbery has increased

from what it was when the average sentence was 7 years. He obtains data on

400 more recent robberies and finds an average time served of 7.5 years. A 95%

confidence interval for the average time served is:

a. (7.25, 7.75).

b. (7.21, 7.79).

c. (7.11, 7.89).

37.5 1.96

400

6.1 Estimating with Confidence

6.1-1

The manager at a movie theater would like to estimate the true mean amount of money spent by customers on popcorn only. He selects a simple random sample of 26 receipts and calculates a 92% confidence interval for true mean to be ($12.45, $23.32). The confidence interval can be interpreted to mean that, in the long run:

a. 92% of similarly constructed intervals would contain the population mean.

b. 92% of similarly constructed intervals would contain the sample mean.

c. 92% of all customers who buy popcorn spend between $12.45 and $23.22.


6.1-2

Lumber intended for building houses and other structures must be monitored for strength. A random sample of 25 specimens of Southern Pine is selected, and the mean strength in calculated to be 3700 pounds per square inch. Strengths are known to follow a normal distribution with standard deviation 500 pounds per square inch. A 90% confidence interval for the true mean strength of Southern Pine is:

a. (3200, 3647).

b. (3402, 3794).

c. (3536, 3865).


6.1-3

3700 (1.645)(50025

)

You want to rent an unfurnished one-bedroom apartment in Boston next year. A 95% confidence interval for the true mean monthly rent of all one bedroom apartments in Boston is: (1263.64, 1536.36). What is the margin of error for this interval?

a. 126.43

b. 136.36

c. 143.46


6.1-4

MOE = ½(length of the interval) = ½(1536.36 – 1263.64)

A student curious about the average number of chocolate chips in a commercial

brand of cookie estimated the standard deviation to be 8. If he wants to be 99%

confident in his results, how many chocolate chip cookies will he need to sample

to estimate the mean to within 2?

a. 11

b. 107

c. 62

2

106.17z

nm


6.1-5

Lumber intended for building houses and other structures must be monitored for strength. Good lumber is known to have a standard deviation of 500 lbs. per square inch. At 80% confidence, what is the required sample size for the margin of error to be +/- 250lbs.?

a. 6

b. 7

c. 8


6.1-6

n z*m

2

We would like to estimate the true mean number of hours adults sleep at night. Suppose that sleep time is known to follow a normal distribution with standard deviation 1.5 hours. What sample size is required in order to estimate the true mean to within 0.5 hours with 96% confidence?

a. 36

b. 37

c. 38


6.1-7

An instructor wanted to construct a confidence interval for the mean

GPA of the students in his class. He used the campus records system

to obtain their GPAs and computed the 95% interval as (2.32, 2.87). If

he wants to use this interval to describe the students in his class:

a. he's 95% confident the interval contains the real average GPA.

b. there's a 5% chance the interval is wrong.

c. he didn't need an interval after all. The mean calculated is exactly


6.1-8

To construct a 92% confidence interval, the correct z* to use is

a. 1.75

b. 1.41

c. 1.645


6.1-9

To construct a 92% confidence interval, the correct z* to use is

a. 1.75

b. 1.41

c. 1.645

Calculator: invNorm(.04) or invNorm(.96)


6.1-9c

C

A radio station wants to know if residents in their area are in favor of a

proposed tax increase. They invite listeners to call in to respond to the

poll. Of the 800 who responded, 645 were against the tax. The station

calculated a 95% confidence interval and declared "Between 77.9%

and 83.4% of residents oppose the new tax."

a. This is a valid interval, so the station must be right.

b. We can't say for certain that between 77.9% and 83.4% of residents

oppose the new tax because it is a 95% confidence interval.

c. Because of the way the poll was conducted, the results are invalid.Voluntary Sample


6.1-10

A 95% confidence interval for the mean, μ, of a population is (13, 20).

Based on this interval:

a. there is a 95% chance μ is in the interval.

b. 95% of the observations lie in the interval.

c. the method gives correct results 95% of the time.


6.1-11

Crop researchers plant 100 plots with a new variety of corn. The

average yield for these plots is = 130 bushels per acre. Assume that

the yield per acre for the new variety of corn follows a normal

distribution with unknown mean µ and standard deviation = 10

bushels per acre. A 90% confidence interval for µ is:

a. 130 ± 1.645.

b. 130 ± 1.96.

c. 130 ± 16.45.

x

101.645

100x


6.1-12

A random sample of size n is collected from a population with standard

deviation . With the data collected, a 95% confidence interval is

computed for the mean of the population.

Which of the following would produce a new confidence interval with

smaller width (smaller margin of error) based on these same data?

a. Increase .

b. Use a smaller confidence level.

c. Use a smaller sample size.


6.1-13

We would like to construct a confidence interval to estimate the mean μ of some variable X. Which of the following combinations of confidence level and sample size will produce the shortest interval?

a. 90% confidence, n = 10

b. 95% confidence, n = 10

c. 95% confidence, n = 20

d. 99% confidence, n = 10

e. 99% confidence, n = 20


6.1-14

A researcher reports that a test is "significant at 5%." This test will be:

a. significant at 1%.

b. not significant at 1%.

c. significant at 10%.

6.2 Tests of Significance

6.2-1

The response times of technicians of a large heating company follow a normal distribution with a standard deviation of 10 minutes. A supervisor suspects that the mean response time has increased from the target of 30 minutes. He takes a random sample of 25 response times and calculates the sample mean response time to be 33.8 minutes. What is the value of the test statistic for the appropriate hypothesis test?

a. Z = 1.65

b. Z = 2.09

c. Z = 0.77

d. Z = 1.90

e. Z = 1.83


6.2-2

z 33.8 3010

25

1.90

A researcher wants to know if the average time in jail for robbery has

increased from what it was several years ago when the average

sentence was 7 years.

He obtains data on 400 more recent robberies and finds an average

time served of 7.5 years. If we assume the standard deviation is 3

years, what is the P-value of the test?

a. 0.0004

b. 0.0008

c. 0.9996

7.5 73.33 ( 3.33)

3

400

z P z

6.2-3


A researcher wants to know if the average time in jail for robbery has

increased from what it was several years ago when the average

sentence was 7 years.

He obtains data on 400 more recent robberies and finds an average

time served of 7.5 years. If we assume the standard deviation is 3

years, what is the P-value of the test?

a. 0.0004

b. 0.0008

c. 0.9996

Calculator: normalcdf(3.33,1000) or normalcdf(7.5,1000,7,3/√400 )

6.2-3c

C

7.5 73.33 ( 3.33)

3

400

z P z


A researcher wants to know if tougher sentencing laws have had a

positive effect in terms of deterring crime. He plans to select a sample

of states which have enacted a "3 strikes" law and compare violent

crime rates before the law was enacted and two years later.

The correct set of hypotheses to test are:

a. H0: and Ha:

b. H0: and Ha:

c. H0: and Ha:

d. H0: and Ha:

before after

6.2-4

before after

before after before after

before afterx x before afterx x

before afterx x before afterx x

before after before after


What is the P-value for a test of the hypotheses H0: μ = 10 against Ha: μ ≠ 10

if the calculated statistic is z = 2.56?

a. 0.0052

b. 0.0104

c. 0.9948

2 ( 2.56)P z

6.2-5


What is the P-value for a test of the hypotheses H0: μ = 10 against Ha: μ ≠ 10

if the calculated statistic is z = 2.56?

a. 0.0052

b. 0.0104

c. 0.9948 Calculator: 2 normalcdf(-1000,-2.56)

6.2-5c

C

2 ( 2.56)P z


The times for untrained rats to run a standard maze has a N (65, 15) distribution

where the times are measured in seconds.

The researchers hope to show that training improves the times.

The alternative hypothesis is:

a. Ha: µ > 65.

b. Ha: > 65.

c. Ha: µ < 65.

d. Ha: < 65.

6.2-6

x

x


Suppose we are testing the null hypotheses H0: µ = 50

Ha: µ ≠ 50

for a normal population with = 6. The 95% confidence(51.3, 54.7). Then:

a. the p -value for the test is greater than 0.05.

b. the p -value for the test is less than 0.05.

c. the p -value for the test could be greater or less than 0.05. It can't be determined

without knowing the sample size.

50 (51.3, 54.7) , 0.05Therefore p value

6.2-7


A sample was taken of the salaries of 20 employees from a largecompany. The following are the salaries (in thousands of dollars) forthis year (the data are ordered).

Suppose each employee in the company receives a $3,000 raise for next year (each employee's salary is increased by $3,000). The interquartile range of the salaries will:

a. be unchanged.

b. increase by $3,000.

c. be multiplied by 3,000.

28 31 34 35 37 41 42 42 42 47 49 51 52 52 60 61 67 72 75 77

6.3 Use and Abuse of Tests

6.3-1

A distribution has a mean of 100 and a median of 120. The shape of

this distribution is most likely:

a. skewed left.

b. skewed right.

c. symmetric.

6.3-2


Which of the following measures are not affected by outliers?

a. The mean

b. The standard deviation

c. The IQR

6.3-3


Does the value of the standard deviation depend on the value of the

mean?

a. Yes. If the mean gets larger the standard deviation will also get larger.

b. Yes. You need to know the mean to be able to calculate the standard

deviation.

c. No. The center of a distribution and the spread are not related.

22 ( )

1ix x

sn

6.3-4


A teacher gave a 25-question multiple choice test. After scoring the tests,

she computed a mean and standard deviation of the scores. The

standard deviation was 0. Based on this information:

a. all the students had the same score.

b. she must have made a mistake.

c. about half the scores were above the mean.

6.4-1

6.4 Power and Inference as a Decision*

The 5-number summary of scores on a test is:

35 60 65 70 90

Based on this information:

a. there are no outliers.

b. there are low outliers.

c. there are both high and low outliers.

1.5 (IQR) = 15

-15 15

6.4-2


The time for students to complete a standardized placement exam

given to college freshman has a normal distribution with a mean of 62

minutes and a standard deviation of 8 minutes. If students are given

one hour to complete the exam, the proportion of students who will

complete the exam is about:

a. 0.25.

b. 0.40.

c. 0.60.

( 0.25)P z

6.4-3


The scores on a university examination are normally distributed with a mean

of 62 and a standard deviation of 11. If the top 10% of students are given an

A, what is the lowest mark that a student can have and still be awarded an

A?

a. 63.28

b. 70.97

c. 76.08 62 + (1.28)(11)

6.4-4


In which of the following situations is a Type II error more serious than a Type I error? Suppose you have to decide whether or not to:

I. Fill your gas tank before crossing a desert

Ho: Will have enough gas

Ha: Will not have enough gas

II. Slow down to the speed limit

Ho: Car following is a police car

Ha: Car following is not a police car

III. Study a topic for an exam

Ho: The topic will be on the exam

Ha: It will not be on the exam

a. I only

b. II only

c. II and III

d. I and II

6.4-5


A box of Teddy Grahams chocolate cookies is supposed to weigh 250 grams. There is some variation in weight from box to box. The box contents are normally distributed with an unknown mean and a known standard deviation of 3 grams. A consumer agency wishes to conduct a 5% significance test to test whether the mean weights differ from 250 grams. A random sample of size 8 is selected; what is the power of the test when the alternative hypothesis is Ha: μ ne 245?

a. 0.0011

b. 0.9374

c. 0.9970

6.4-6


reject the null when x-bar < 247.9 or x-bar > 252.08

Inference and Confidence Intervals

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