B Heard

Not to be used, posted, etc. without my expressed permission. B Heard

The final exam is cumulative.

It should take you around the full amount of time given.

Make sure you check the number of pages, etc. before starting.

Not to be used, posted, etc. without my expressed permission. B Heard

What you should know……….

Not to be used, posted, etc. without my expressed permission. B Heard

Correlation Be able to identify types of correlation looking at a

scatter plot (weak, moderate, strong, positive/negative).

Remember that the correlation coefficient is between -1 and 1 and we use “r” to represent it (NOT r^2)

Using pivot tables to calculate probabilities

Be able to use a pivot table to calculate probabilities. For example

First Graders Ninth Graders Sr. Citizens Raw Totals

Vanilla 10 12 16 38

Chocolate 34 37 14 85

Strawberry 6 11 10 27

Raw Totals 50 60 40 150

First Graders Ninth Graders Sr. Citizens Raw Totals

Vanilla 10 12 16 38

Chocolate 34 37 14 85

Strawberry 6 11 10 27

Raw Totals 50 60 40 150

This is a survey of First Graders, Ninth Graders, and Senior Citizens

(all together) and their favorite ice cream of the three choices.

What are the following?

P(Ninth Grader) = 60/150 = 6/15 = 2/5

P(Chocolate) = 85/150 = 17/30

P(Sr. Citizen and prefer Vanilla) = 16/150 = 8/75

P(First Grader given that they prefer Vanilla) = 10/38 = 5/19

P(Prefer Strawberry given that they are a Ninth Grader) =

11/60

First Graders Ninth Graders Sr. Citizens Raw Totals

Vanilla 10 12 16 38

Chocolate 34 37 14 85

Strawberry 6 11 10 27

Raw Totals 50 60 40 150

Be able to understand the normal distribution and how it relates to the mean, standard deviation, variance, etc.

For example, I did an analysis and found the mean number of failures was 7 and the standard deviation was 1.5. Answer the two questions below.

How many standard deviations is 10 from the mean? 10 – 7 = 3, 3/1.5 = 2 (your answer)

How many standard deviations is 6.25 from the mean? 6.25 – 7 = - .75, - .75/1.5 = -0.5 (your answer)

Be able to use the Standard Normal Distribution Tables or

Excel to find probability values and z scores.

Examples:

Find the following probability involving the Standard Normal Distribution. What is P(z<1.55)?

.9394 (from the table or use Excel “=NORMDIST(1.55,0,1,TRUE)”)

Find the following probability involving the Standard Normal Distribution. What is P(z > -.60)?

1 – 0.2743 = 0.7257 (using table or Excel 1 – “=NORMDIST(-0.6,0,1,TRUE)”)

A researcher is performing a hypothesis test on a claim about a population proportion. Using an alpha = .04 and n = 80, what is the rejection region if the alternate hypothesis is Ha: p > 0.70?

Alternate hypothesis test shows that this is a Right

Tailed test (since it’s p > 0.70) with a right tail area of .04

(since alpha = .04). Therefore we are going to reject Ho

if z > 1.75 (looked in standard normal table to find z

score for a probability of 0.96, z of 1.75 was the closest)

A researcher is performing a hypothesis test on a claim about a population proportion. Using an alpha = .03 and n = 95, what two critical values determine the rejection region if the null hypothesis is: Ho: p = 0.44?

Since Ho: p = 0.44, this is a two tailed test. Each tail has

an area = .03/2 = .015. The z-values that correspond to

this area in the tail is +/- 2.17. (You can see this by

finding the z score for either .015 or .985 realizing it’s two

tailed)

A manufacturer claims that the mean lifetime of its computer component is 1100 hours. A buyer’s researcher selects 49 of these components and finds the mean lifetime to be 1105 hours with a standard deviation of 30 hours. Test the manufacturer's claim. Use alpha = .02.

Answer on following chart

Ho: mu = 1100 hours (claim);

Ha: mu does not = 1100 hours; two tailed test, therefore, .01 is in left tail

and .01 is in right tail; thus critical values are ± 2.33;

test statistic is

z = (xbar - mu) ÷ [sigma ÷ sqrt(n)] = (1105 - 1100) ÷ [30 ÷ sqrt(49)] = 5 ÷ [30 ÷ 7] = 5 ÷ 4.29 = 1.17 which is in the do not reject area because p value corresponding to z= +1.17 is 0.879

Fail to reject Ho. (Because 1.17 is in the bounds of the critical values ±

2.33)There is not enough evidence to reject the manufacturer's claim that

the mean lifetime is 1100 hours.

A Pizza Delivery Service claims that it will get its pizzas delivered in less than 30 minutes. A random selection of 49 service times was collected, and their mean was calculated to be 28.6 minutes. The standard deviation was 4.7 minutes. Is there enough evidence to support the claim at alpha = .10. Perform an appropriate hypothesis test, showing each important step. (Note: 1st Step: Write Ho and Ha; 2nd Step: Determine Rejection Region; etc.)

Answer following chart

Ho: mu >= 30 min.

Ha: mu < 30 min. (claim). Therefore, it is a left-tailed test.

n=49; x-bar=28.6; s=4.7; alpha=0.10

Since alpha = 0.10, then the critical z value will be zc = -1.28

since n>30 then s can be used in place of sigma.

Standardized test statistic z = (x-bar - mu)/(s/sqrt(n)) z = (28.6-30)/(4.7/sqrt(49))z = -2.085

since -2.085 < -1.28, we REJECT Ho.

That is, at alpha = 0.10, There is enough evidence to support the

Pizza Delivery Service’s claim.

(p-value method could have also been used)

Determine the minimum required sample size if you want to be 90% confident that the sample mean is within 5 units of the population mean given sigma = 8.4. Assume the population is normally distributed.

n = (Zc*sigma/E)^2 = [(1.645 * 8.4)/ 5]^2 = (2.7636)^2= 7.64= 8 (always round up sample sizes)

A researcher wants to estimate the true proportion of failures caused my a minimum of one event. The researcher wants to be within 5% of the true proportion when using a 95% confidence interval. A previous study estimated the population proportion at 0.64.

(a). Using this previous study as an estimate for p, what sample size should be used?

(b). If the previous study was not available, what estimate for p should be used?

Answer follows

(a). The critical z-value for a 95% confidence interval is

1.96. Since a previous study is known, we can use it to

estimate p = 0.64. The maximum error is 0.05.

Sample size = n = p*(1-p)*( z / error )^2= 0.64*(.36)*(1.96/0.05)^2= 354.04 = 355 rounded up

Thus, at least 355 failures must be sampled.

(b). If no estimate of p is known, we must use p = 0.5 to

have a large enough sample size to meet the desired

maximum error. (i.e. use p = 0.5 when you don’t know it)

Variance/Standard Deviation… how they relate

If the variance of some data is 91, what is the standard deviation?

Square root of 91 = 9.54

If the standard deviation of some data is 17, what is the variance?

17^2 = 289

The failure times of a component are listed in hours. {100, 95, 120, 190, 200, 200,280}.Find the mean, median, mode, variance, and range.Do you think this sample might have come from a normal population? Why or why not?

mean = 169.3

median =190

mode = 200

variance = 4553.6

range = 185

Doubtful it came from a normal, compare mean, median,

mode, etc.

The random variable X represents the annual salaries in dollars for a group of entry level accountants. Find the expected value E(X).X = {$30,000; $38,000; $42,000}.P(30,000) = .2; P(38,000) = .7; P(42,000) = .1

E(X) = 30000*.2 + 38000*.7 + 42000*.1 = $36,800

The average (mean) monthly grocery cost for a family of 4 is $600. The distribution is known to be “normal” with a standard deviation = 60. A family is chosen at random.

a) Find the probability that the family’s monthly grocery cost purchases will be between $550 and $650.

b)Find the probability that the family’s monthly grocery cost purchases will be less than $700.

c) What is the probability that the family’s monthly grocery cost purchases will be more than $630?

Answers follow

Using Tables or EXCEL (I used an Excel Template like one I showed you):a) P(550 < x < 650) = 0.5953b) P(x < 700) =.9522c) P(x > 630) = .3085

The earnings per share (in dollars) for The Very Pretty Products Company are given by the equation

y-hat = 0.863 + 0.029a - 0.011b

where "a" represents total revenue (in billions of dollars) and "b" represents total net worth (in billions of dollars). Predict the earnings per share when total revenue is $6 billion and net worth is $1 billion.

y-hat = 0.863 + 0.029a - 0.011b

y-hat = 0.863 + 0.029*6 - 0.011*1

y-hat = 1.026

The time required to produce a product was normally distributed with a mean 10.5 days and a standard deviation of 1.5 days (i.e., 36 hours). What is the probability that it will take more that 11 days to produce the product?

We want P(X > 11)z = (x - mu)/sigma= (11 – 10.5)/1.5 = .5/1.5 = .33P(z > .33) = 1 - P(z < .33)= 1 – 0.6293= 0.3707

A shipment of 30 Widgit’s contains 7 defective Widgits. How many ways can a Widget buying company buy 3 of these units and receive no defective units?

There are 23 Widgets which are not defective.Thus there are 23C3 ways to get 3 sets of Widgets with none defective.23C3 = 1771 (using “combin” function in Excel)

For the following statement, write the null hypothesis and the alternative hypothesis. Then, label the one that is the claim being made.A car manufacturer claims that the mean life time of a car is more than 7 years.

Ho: mu <= 7 yearsHa: mu > 7 years (claim)

Remember that Ho always contains equality and the claim can be either Ho or Ha

A State Trooper notes that at a certain intersection, an average of three cars run the red-light per hour. What is the probability that the next time he is there exactly two cars run the red-light?

Poisson with average of 3. want P(2) P(2) = .2240 (Use Excel or Excel Template)

The probability that a house in a neighborhood has a dog is 40%. If 50 houses in the neighborhood are randomly selected what is the probability that one (or a certain number) of the houses will have a dog?

a. Is this a binomial experiment?

b. Use the correct formula to find the probability that, out of 50 houses, exactly 22 of the houses will have dogs. Show your calculations or explain how you found the probability.

Answer Follows

a) Fixed number of independent trials, only two possible outcomes in each trial {S,F} (dog or not), probability of success is the same for each trial, and random variable x counts the number of successful trials. So YES it is.

b) n = 50; p = .40 = P(success) = house has a dogWe want P(22) --> P(22) = .0959 (I used a

Binomial Template)

The stem and leaf plot for the following data is displayed below: {11,11,12,16,17,24,25,26,26,34,36,37,44,46,51,53,62}

Stem and Leaf Plot:1|112672|45663|4674|465|136|2

Discuss the shape of the data distribution.

It is right skewed. (other choices would be left skewed, symmetric, etc). This one is right skewed because if you turned it 90 degrees counter clockwise the “tail” would be on the right.

Scores on an exam for entering a military training program are normally distributed, with a mean of 60 and a standard deviation of 12. To be eligible to enter, a person must score in the top 15%. What is the lowest score you can earn and still be eligible to enter?

mu = 60; sigma = 12we want top 15% or an area greater than 1 - .15 or .85z = 1.04 ---> x = (1.04)(12) + 60 = 72.48 orneed a score of 73 (Round it up)

A polling company wants to estimate the average amount of contributions to their candidate. For a sample of 100 randomly selected contributors, the mean contribution was $50 and the standard deviation was $8.50.

(a) Find a 95% confidence interval for the mean amount given to the candidate

(b) Interpret this confidence interval and write a sentence that explains it.

Answer Follows

(a). Since sample size = n = 100> 30, we can use a z-value. For a 95% confidence level, z-value = 1.96. Also, sample mean = xbar = 50; population standard deviation is estimated by sample standard deviation (since n > 30) = s = 8.50

E = z * s / sqrt(n) = 1.96 * 8.50/sqrt(100) = 1.666

xbar + E = 50.00 + 1.67 = 51.67xbar - E = 50.00 - 1.67 = 48.33

Thus, 95% confidence interval = ($48.33,$51.67)

(b) We are 95% confident that the population mean amount contributed is between $48.33 and $51.67

Private Johnson earned a 78 on his history test and 82 on his math test. In the history class the mean score was 79 with standard deviation of 6. In the math class the mean score was 84 with standard deviation of 5.

Convert each score to a standard score (Z score)

Which score was the higher with respect to the rest of the class? Show your work and explain your answer.

history: mu = 79; sigma = 6; z = -0.167 (78-79/6)math: mu = 84; sigma = 5; z = -0.400 (82-84/5)thus an 78 in history is better than an 82 in math.

(because -0.167 is greater than -0.400)