# Week8 Live Lecture for Final Exam

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04-Jul-2015Category

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### Transcript of Week8 Live Lecture for Final Exam

- 1. B Heard

Lecture for theFinal ExamStatistics For Decision Making

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

2.

- The final exam is cumulative. 3. It should take you around the full amount of time given. 4. Make sure you check the number of pages, etc. before starting.

Final Exam Review

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

5. Final Exam Review

What you should know.

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

6. 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)

Final Exam Review

7. Using pivot tables to calculate probabilities

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

Final Exam Review

8. Final Exam Review

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

9. Final Exam Review

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

10. 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)

Final Exam Review

11. 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 -.60)?

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

Final Exam Review

12. A researcher is performing a hypothesis teston a claim about a population proportion. Usingan alpha = .04 and n = 80, what is the rejection region if the alternate hypothesisis Ha: p > 0.70?

Alternate hypothesis test shows that this is a Right

Tailed test (since its 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)

Final Exam Review

13. A researcher is performing a hypothesis teston a claim about a population proportion. Usingan alpha = .03 and n = 95, what two critical values determine the rejection regionif 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 its two

tailed)

Final Exam Review

14. A manufacturer claims that the mean lifetime of its computer component is 1100 hours. A buyers 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

Final Exam Review

15. 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 = 1.17which is in the do not reject area because p value corresponding to z= +1.17 is 0.879

Fail toreject 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.

Final Exam Review

16. A Pizza Delivery Service claims thatit 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:WriteHo and Ha; 2nd Step: Determine Rejection Region; etc.)

Answer following chart

Final Exam Review

17. 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 Services claim.

(p-value method could have also been used)

Final Exam Review

18. 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)

Final Exam Review

19. 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

Final Exam Review

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

1.96. Since a previous study is known, we canuse 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 = 0.5to

have a large enough sample size to meet the desired

maximum error. (i.e. use p = 0.5 when you dont know it)

Final Exam Review

21. 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

Final Exam Review

22. 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.

Final Exam Review

23. The random variable X represents the annualsalaries in dollarsfor 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

Final Exam Review

24. 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 familys monthly grocery cost purchases will be between $550 and $650.b)Find the probability that the familys monthly grocery cost purchases will be less than $700.c) What is the probability that the familys monthly grocery cost purchases will be more than$630?

Answers follow

Final Exam Review

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

Final Exam Review

26. 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 billionand 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

Final Exam Review

27. The time required to produce a product was normally distributed with a mean10.5 days and a standard deviation of1.5 days (i.e., 36hours).What is the probability that it will take more that11days 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

Final Exam Review

28. A shipment of 30 Widgitscontains7 defective Widgits.How many ways can a Widget buying company buy3 of these units and receive no defective units?

There are 23 Widgets which are not defective.Thus there are 23C3ways to get3 sets of Widgets with none defective.23C3 = 1771 (using combin function in Excel)

Final Exam Review

29. 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 t