252y0621 11/8/06

ECO252 QBA2Name KEY

SECOND HOUR EXAMCircle Hourof Class Registered

November 8, 2005MWF2, MWF3, TR12:30, TR2

Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not usually acceptable.

I. (8 points) Do all the following.Make diagrams!

- If you are not using the supplement table, make sure that I know it.

1.

For make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area between -1.11 and 1.67. Because this is on both sides of zero we must add togrther the area between -1.11 and zero and the area between zero and 1.67. If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 20. Indicate the mean by a vertical line! Shade the area between 10 and 35. This area includes the mean (20), and areas to either side of it so we add together these two areas.

2.

For make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the entire area above -1.89. Because this is on both sides of zero we must add togrther the area between -1.89 and zero and the entire area above zero.If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 20. Indicate the mean by a vertical line! Shade the entire area above 3. This area includes the mean (20), and areas to either side of it so we add together these two areas.

3.

For make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area between -2.22 and zero. Because this is completely on the left of zero and touches zero, we can simply look up our answer on the standardized Normal table. If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 20. Indicate the mean by a vertical line! Shade the area between 0 and 20. This area includes the mean (20), but does not include any points to the right of the mean, so that we neither add nor subtract.

4. (Do not try to use the t table to get this.) For make a diagram. Draw a Normal curve with a mean at 0. is the value of with 5.5% of the distribution above it. Since 100 – 5.5 = 94.5, it is also the 94.5th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should show that the probability between and zero is 94.5% - 50% = 44.5% or The closest we can come to this is (1.61 is also acceptable here.) So To get from to , use the formula , which is the opposite of . . If you wish, make a completely separate diagram for . Draw a Normal curve with a mean at 20. Show that 50% of the distribution is below the mean (7). If 5.5% of the distribution is above , it must be above the mean and have 44.5% of the distribution between it and the mean.

Check:

II. (22+ points) Do all the following? (2points each unless noted otherwise). Look them over first. The exam is normed on 50 points. The computer problem is at the end.

  1. A company gives an exam to graduates of quality control programs in two plants.Samples of scores are as follows:

Boston 90 73 78 82 66

Atlanta 81 72 50 66 55 70

a. Compute the sample variance for Boston – Show your work! The sample mean and standard deviation for Atlanta are 65.67 and 11.43. (2)

b. Is there a significant difference between the scores in the two plants? You may assume that variances are equal. State your hypotheses! (2)

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252y0621 11/8/06

Solution:

Row

1 90 8100 12.2 148.84

2 73 5329 -4.8 23.04

3 78 6084 0.2 0.04

4 82 6724 4.2 17.64

5 66 4356 -11.8 139.24

389 30593 0.0 328.80

,,

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252y0621 11/8/06

The formula table gives us the formulas below. (Method D2 – Comparison of two means with samples coming from populations with similar variances.)

Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical Value
Difference
between Two
Means (
unknown,
variances
assumed equal) /

/ /
/

, , , and

Our hypotheses are or or if , . .

. This is the pooled variance. . so .

. Recall that our alternate hypothesis is so this is a two-sided test. You may have said

Test Ratio:.

If this test ratio lies between , do not reject . . Make a diagram with zero in the middle showing shaded ‘reject’ regions below -2.262 and above 2.262. Since –1.918 does not fall in the 'reject' region, do not reject . Or you can say that, since 1.918 falls between and , for a one-sided test, . But this is a two-sided test so that Since the p-value is above , do not reject .

Critical Value: . For a two-sided test we want two critical values above and below If is between the critical values, do not reject . . Make a diagram with 0 in the middle showing shaded ‘reject’ regions below -14.307 and above 14.307. Since does not fall in the 'reject' region, do not reject .

Confidence Interval:. We already know that so we can say or .Make a diagram with 12.13 in the middle. Represent the confidence interval by shading the area between -2.18 and 26.44. Since zero is in this area, do not reject . Or simply note that the error part of the confidence interval is larger that the sample mean difference so the interval must include zero.

Exhibit 1

The director of the MBA program of a state university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. A random sample of 215 potential students who have not taken the orientation is compared with a random sample of 215 students who have taken the orientation. Of the first group 130 students viewed the program as ‘good.’ Of group that had taken the orientation 164 viewed the program as ‘good.’ She wishes to see if the fraction that considered the program ‘good’ has risen because of the orientation.

  1. Referring to Exhibit 1, which test should she use? (2)

a)-test for difference in proportions

b)*z-test for difference in proportions

c)McNemar test for difference in proportions

d)Wilcoxon rank sum test

  1. Referring to Exhibit 1, what is the null hypothesis? (2)

Solution: She wants to know if which is the same as . So we have or

  1. Referring to Exhibit 1, what is the value of the computed test statistic? (4)[12]

Solution: or or if , are implied. If you use we have

For the first definition

  1. Referring to Table Exhibit 1, what is the p-value of the test statistic?(2)[14]

Solution: The approximate p-value will be, since this is a left sided test

Exhibit 2

(Dummeldinger) A researcher took a random sample of graduates of MBA programs, which included women and men. Their starting salaries were recorded. Use and/or for population parameters for women and and/or for men. The sample yields following data. and . . The researcher wants to show that men have a higher mean starting salary than women.Assume that these are independent samples

Hint: To preserve both our sanity, move the decimal point three places to the right in both the means and standard deviations. You will be working in thousands but will get the same value for the test ratio.

We want to see if the means or medians, as appropriate, are different.

Assume that these are independent samples from population with a Normal distribution and that .

  1. Referring to Exhibit 2, if what is the null hypothesis? (1)

a)

b)

c)

d)*

e)

f)None of the above. (Give the correct one!)

Explanation: It says ‘men have a higher mean starting salary than women.’ This is or . Because this does not contain an equality, it must be an alternate hypothesis!

  1. Referring to Exhibit 2, if we do not have a computer available, which of the following methods would be most practical (and correct) for you to use? (2)

a)*z- test comparing two means

b)t- test comparing two means assuming equal variances

c)t- test comparing two means not assuming equal variances

d)t- test comparing two means for paired data

e)z- test comparing two proportions

f)None of the above. (Give the correct one!)[17]

  1. Referring to Exhibit 2, assume that the correct alternate formula for a critical value is , where t can be replaced by zfor one method. What is the valueof ? (3) [20]

Solution: From the formula table. We can use this with replacing for large samples.

Interval for / Confidence
Interval / Hypotheses / Test Ratio / Critical Value
Difference
between Two
Means (
known) / / / /

=1.4656.

  1. Referring to Exhibit 2, and the previous problems, what is the test ratio that we would use to test your hypothesis? (2)

or where and .

so .

  1. Referring to Exhibit 2, and the previous problems, assuming that your null hypothesis is correct, do you reject the null hypothesis? Why? (2) [24]

Solution:Recall that our hypotheses were this is a left-sided test. so , . If we use a test ratio the rejection region is below -1.282. Since our value of z is not in the rejection region, do not reject the null hypothesis! If you use a critical value, you want a critical value below zero so becomes and our rejection region is below thia point. Since is not in the rejection zone do not reject the null hypothesis. The Minitab run follows.

MTB > TwoT 150 48266.7 13577.63 150 50100 11741.25;

SUBC> Alternative -1.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 150 48267 13578 1109

2 150 50100 11741 959

Difference = mu (1) - mu (2)

Estimate for difference: -1833.30

95% upper bound for difference: 585.14

T-Test of difference = 0 (vs <): T-Value = -1.25 P-Value = 0.106 DF = 291

  1. (Extra credit) compute a two-sided confidence interval for the ratio of the two variances in the previous problem. (3)

Solution:The formulas given in the first article in the syllabus supplement is or

and . so and . Because we need . This should be close to or or maybe or . It looks like maybe the best guess would be abovt halfway be tween the first two like 1.33 so the intervals are or

Exhibit 3

You drive a New York avenue with 25 traffic lights on it. You suspect that this is equivalent to playing a game where the probability of success (getting a red light) is 50% (There are 25 tries.). You use a binomial table to figure out the probabilities of getting various numbers of red lights on a single run and then you record the number of times you get these numbers of red lights during 25 trips up the avenue.

The left column is copied from my cumulative binomial table. The next column comes from differencing the first column. The third column is the second column multiplied by 25. The fourth column gives the numbers actually observed. The last two columns show the process of making the observed data into a cumulative distribution.

EO Cum O Cum

0.11476 0.11476 2.86903 3.120

0.21218 0.09742 2.43554 7.280

0.34502 0.13284 3.3210310.400

0.50000 0.15498 3.8745212.480

0.65498 0.15498 3.8745315.600

0.78782 0.13284 3.3210318.720

0.88524 0.09742 2.4355422.880

1.00000 0.11476 2.86903251.00

  1. Referring to Exhibit 3. The correct way to test for the distribution cited is: (2)

a)*Kolmogorov-Smirnoff Test

b)Lilliefors Test

c)Chi-squared test with 8 degrees of freedom

d)Chi-squared test with 7 degrees of freedom

e)Chi-squared test with 6 degrees of freedom

f)test for a proportion

g)None of the above. (Give the correct one!)[26]

Comment: KS is preferred for small samples where the parameters are not estimated from the data. Notice that the Es are too small to use chi-squared.

.

  1. Referring to Exhibit 3. Assume that your choice of method in the previous problem is correct. Carry out the test. (4) [30]

Solution:

We can copy a cumulative and E from the table.

D

0.11476 .120 .00524

0.21218 .280 .06782

0.34502 .400 .05498

0.50000 .480 .02000

0.65498 .600 .05498

0.78782 .720 .06782

0.88524 .880 .00524

1.00000 1.00 0

The table gives a 5% critical value that is .165 for None of s is that big!.

  1. (Abramovic) You are producing a subassembly to be delivered to another firm. Historically 85% of the assemblies have come off the assembly line good and ready to deliver. 10% have come out defective, but repairable. The remaining 5% have to be trashed. You take two independent samples of 200 subassemblies, one before and one after a ‘zero defects’ program. is the data from the first sample. is the data from the second sample.

Good150160

Repairable 35 37

Trashed 15 3

a) Using the historical percents in the problem in a chi-squared test put together an E column. (1)

b) Do the chi-squared test to compare and E. (3)

c) Do a confidence interval for , the amount that the proportion of ‘good’ items has improved because of the ‘zero defects’ program. (3) [37]

d) Make an 89% confidence interval for the same difference. Do not use the t table.

Solution:

a) I got Eby multiplying the historical proportions by 200.

Row

1 150 0.85 170 -20 2.3529 132.353

2 35 0.10 20 15 11.2500 61.250

3 15 0.05 10 5 2.5000 22.500

200 1.00 200 0 16.1029 216.103

b)

or . Both of these two formulas are shown above. There is no reason to do both.. . So we have or . Since we reject the null hypothesis.

c) The Formula Table says the following.

Interval for / Confidence
Interval / Hypotheses / Test Ratio / Critical Value
Difference
between
proportions
/

/
/

Or use /

, , and

, , and

If a 95% confidence interval would be

Does this mean that there has been no significant improvement?

d) Make an 89% confidence interval for the same difference. Do not use the t table.Solution:This is the easiest problem on the exam. All you need to know is , which you found on page 1.

  1. (Abramovic) The following average Mathematics SAT scores are classified by gender and year.

(Compare with question on 252y0622)

Year Female Male

1467514

2470512

3470470

4465409

5466507

6461505

7460502

8459501

9449495

10446497

If we wish to show that male results exceed female results and suspect that the underlying distribution is not Normal. We should use: (2)

a)Sign Test (Actually possible, but not powerful enough)

b)Wilcoxon-Mann-Whitney Test for independent samples

c)*Wilcoxon signed rank test

d)t-test for paired data

e)t-test for independent samples with equal variances.

f)F-test.

g)None of the above. (Give the correct one!)

  1. Turn in your computer output for the first problem only. To get full credit it must be noted what hypotheses were tested and what were the results. Use a 5% significance level. (4) [41]

ECO252 QBA2

SECOND EXAM

November 7-8

TAKE HOME SECTION

-

Name: ____KEY______

Student Number: ______

Class days and hour: ______

III. Do at least 3 problems (at least 7 each) (or do sections adding to at least 20 points - Anything extra you do helps, and grades wrap around) . Note: Look at 252thngs (252thngs) on the syllabus supplement part of the website before you start (and before you take exams). Show your work! State and where appropriate. You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated your conclusion.. (Use a 95% confidence level unless another level is specified.) Answers without reasons usually are not acceptable. Neatness and clarity of explanation are expected. This must be turned in when you take the in-class exam. Note that from now on neatness means paper neatly trimmed on the left side if it has been torn, multiple pages stapled and paper written on only one side.

  1. A bank is testing various ways to serve walk-in customers. It tries 10 different new ways to organize this service and gets the following results. Each result can be considered as a random sample of 14 times that it takes a teller to serve ten walk-in customers. The data is presented below. Method 1 is the method currently in use. You are asked to compare method 1 with one other method by comparing the mean or median time it takes to serve the customers and to find out if there is a significant difference between the methods. On the basis of your results you can conclude that i) the old method is better, ii) the new method is better or iii) more information is needed to make a decision. The second method to compare to method 1 is determined by the second to last digit of your student number: if it is 0 use method 10; if it is 1 use method 11. If it is 2, 3, 4 etc use the method with the corresponding number. Label this problem clearly with the number of the method that you are comparing with Method 1.

Method

Row 1 2 3 4 5 6 7 8 9 10 11

1 2.9 2.8 2.6 7.7 2.4 6.6 3.5 3.4 3.5 2.3 3.4

2 3.9 2.6 2.7 2.9 13.4 3.7 8.4 8.3 3.4 6.9 4.4

3 2.6 2.6 3.2 4.3 5.8 9.7 4.3 4.2 3.8 3.3 3.1

4 3.1 2.9 2.8 2.7 1.5 1.9 3.3 3.2 3.4 5.3 3.6

5 3.9 2.9 3.6 3.4 9.8 10.1 11.9 11.0 3.6 3.0 4.4

6 2.6 2.8 2.1 4.4 2.7 4.5 3.7 3.6 3.5 3.3 3.1

7 3.3 2.3 2.3 5.5 2.7 2.9 3.0 2.9 4.8 6.1 3.8

8 3.0 2.4 2.6 3.4 4.5 9.9 2.9 2.8 3.5 3.1 3.5

9 3.5 2.0 2.6 3.4 2.3 3.0 3.6 3.5 5.3 2.6 4.0

10 3.1 2.5 2.9 3.5 5.8 31.5 5.4 5.3 3.7 4.4 3.6

11 3.2 2.4 2.4 4.0 4.8 3.5 4.4 4.3 3.4 15.0 3.7

12 2.4 2.0 2.3 3.4 4.2 5.3 3.0 2.9 3.6 6.9 2.9

13 4.0 4.1 0.5 4.1 5.8 9.8 4.3 4.2 3.8 2.1 4.5

14 4.3 4.3 4.7 3.8 6.1 5.3 5.4 5.3 3.8 10.4 4.8

Minitab gives us the following computations for Method 1.14, 3.271, 0.155, 0.580, 2.400 2.825, 3.150, 3.900, 4.300. For other information, see the end of this document. You can assume that the data for method 1 has an underlying Normal distribution. Use a 95% confidence level.

a) Compute a standard deviation for the method that you are comparing with Method 1. Carry at least as many significant figures as you see in the Minitab computations above. (1)

b) Test for a significant difference between the methods on the assumption that your method represents data taken from the Normal distribution and the times have approximately equal variances. Use a test ratio, critical value or a confidence interval (3) or all three (6).

c) Assume that the data represents a Normally distributed population but the variances are not equal (4 – extra credit)

d) Assume that the Normal distribution does not apply. (4)[11]

e) Test to see if the data from your method is Normally distributed (3)

f) Test to see if the standard deviations of the two methods are equal (1)

g) On the basis of the sections of this problem that you have been able to do write a recommendation. (1) [16]

  1. A researcher asked a group of artists and non- artists whether they believed in extra-sensory perception. The data assembled is as below.

Of 114 artists, 58.77% believed in extrasensory perception. 35.96% more or less believed in extrasensory perception and the remainder did not believe.

Of 344 non artists, 37.50% believed in extrasensory perception, 53.20% more or less believed and the remainder did not believe.

a) Is there any association between being an artist and believing in ESP? (6)