I. (14 Points) Do All the Following. ( Make Diagrams )

252y0121 3/26/01 ECO252 QBA2 Name

SECOND HOUR EXAM Hour of Class Registered (Circle) March 20, 2001 MWF 10 11 TR 12:30 2:00

I. (14 points) Do all the following. (Make diagrams!!!)

5. (The Cumulative probability up to 2) .

6. A symmetrical interval about the mean with 79% probability. We want two

points , so that. Make a diagram showing

6 in the middle at the center of a 79% region split into two areas with probabilities

of .3950. From the diagram, if we replace x by z, .

The closest we can come is or . Since

neither of these is much closer than the other, use , and , or -5.295 to 17.295. To check this note that

7. We want a point , so that. Make a diagram of

showing zero in the middle, .4920 between 0 and and .008 above . From

the diagram, if we replace x by z, The Normal table says

. So , and To check this note

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II. (6 points-2 point penalty for not trying part a.) Show your work!

A test shopper goes to Miller's Supermarkets 8 times and to Albert's Supermarkets 8 times. The amount that was spent using a standardized shopping list is shown below. You can regard these data as two independent random samples from populations with a normal distribution. For Miller's, the sample mean is 115.25 and the sample standard deviation is 1.75255.

Obs Miller's Albert's

1 112 119

2 115 121

3 115 122

4 117 120

5 117 122

6 117 124

7 115 122

8 114 122

a. Compute , the standard deviation for Albert's. (3)

b. Compute a 99% confidence interval for the difference between the population means and assuming that the variances are equal for the two parent populations. According to your confidence interval, is there a significant difference between the population means (You must tell why!)? (3)

c. (Extra Credit) (i) Redo the confidence interval on the assumption that the variances are not equal. (6)

(ii) Use an F-test to tell whether you should have assumed that the variances were or were not equal. (2)

Solution: a) Use for Albert's and for Miller's. and , so .

Line

1 119 14161

2 121 14641

3 122 14884

4 120 14400

5 122 14884

6 124 15376

7 122 14884

8 122 14884

972 118114

b) From Table 3 of the Syllabus Supplement:

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

/ /
/

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Confidence Interval:or 3.81 to 8.69. The

interval does not include 0, so there is a significant difference between the means.

Formally, our hypotheses are or or We reject.

Solution: b) From the formula table;

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

, so use 13 degrees of freedom.

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Confidence Interval: The 2-sided interval is 2.01 or 4.24 to 8.26

c) F test. . To test this at the 1% significance level, test and against . is not on our table but is, and must be larger than 6.99. Since neither ratio is larger than , do not reject Conclude that we should have assumed that the variances were equal.

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III. Do at least 3 of the following 4 Problems (at least 10 each) (or do sections adding to at least 30 points - Anything extra you do helps, and grades wrap around) . You must do problem 1a! Show your work! State and where applicable. Do not answer a question 'yes' or 'no' without citing a statistical test. Use a 95% confidence level unless another level is specified.

1a. Turn in your computer output from computer problem 1 only. (3 - 2 point penalty for not handing this in.)

b. (Bowerman et. al.) A researcher knows that last year's average credit card interest rate was 18.3% but believes that, due to increased competition, current rates are better. She checks the interest rates on a random sample of credit cards and comes up with the data given as 'int' on the next page. What does the first test tell us about whether rates have improved? State the null and alternative hypotheses and, using a 5% significance level, tell whether the researcher is right. (3)

c. She now wishes to compare these rates with rates she finds in a search of banks that advertise on the net. She assembles another random sample as 'int1' and runs a second test. Again state the hypotheses and conclusion. (2)

d. Using the statistics that have been computed for you, verify the value of t in the second test and show graphically how the computer got the p-value. (3)

Data Display

int

15.6 17.8 14.6 17.3 18.7 15.3 16.4 18.4 17.6 14.0 19.2

15.8 18.1 16.6 17.0

MTB > print 'int1'

Data Display

int1

18.5 18.7 13.6 12.3 12.9 11.4 10.2 10.5 17.0 14.1 18.3

17.3 15.1 22.1 14.1 15.7 15.3 15.3 20.5 16.5

MTB > ttest mu=18.3 'int';

SUBC> alt=-1.

First Test

T-Test of the Mean

Test of mu = 18.300 vs mu < 18.300

Variable N Mean StDev SE Mean T P-Value

int 15 16.827 1.538 0.397 -3.71 0.0012

b) Solution: Since the p-value is below reject the null hypothesis of equality and conclude that now interest rates are better (lower).

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Second Test

MTB > twosample 'int''int1';

SUBC> alt=1.

Two Sample T-Test and Confidence Interval

Twosample T for int vs int1

N Mean StDev SE Mean

int 15 16.83 1.54 0.40

int1 20 15.47 3.22 0.72

95% C.I. for mu int - mu int1: ( -0.33, 3.04)

T-Test mu int = mu int1 (vs >): T= 1.65 P=0.055 DF= 28

c) Solution: If is the mean interest rate from the old population and is the rate from the net population, or . Since the p-value is above do not reject the null hypothesis of equality. Conclude that now interest rates on the net cannot be proved to be better (lower).

d) Solution: If we are comparing two samples, from the formula table;

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

Make a diagram showing an almost-normal curve with a mean at zero. The p-value is the probability that , so shade the area above 1.65 and label it 5.5%.

MTB >

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2. Bowerman, O'Connell and Hand tell us that (i) of 30 investors who invested in a bond fund, 7 were dissatisfied, (ii) of 30 investors who invested in a stock fund, 6 were dissatisfied and (iii) of 40 investors who invested in a tax-deferred annuity, 10 were dissatisfied. You may assume that those who were not dissatisfied were satisfied. Use a 99% confidence level.

a. Do a two sided confidence interval for the difference between the proportion satisfied with the bond fund and with the stock fund. (3)

b. Test that the proportion of investors who were satisfied with the stock fund was higher than the proportion who invested in the bond fund. (3)

c. Test the hypothesis that the proportioned satisfied was the same for each investment (6)

Solution: To summarize the information in the problem - and

Investment / Bond Fund / Stock Fund / Annuity
Satisfied / 23 / 24 / 30
Dissatisfied / 7 / 6 / 10
Total / 30 / 30 / 40
Proportion satisfied / .7667 / .8000 / .7500

We are comparing and

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

/
/

Or use /

, , Note that and that and are between 0 and 1.

a) or -.307 to .241. 2/3 of you were following the old exam so slavishly that you never realized that I had asked for a confidence interval!

b) Same as or (Only one of the following methods is needed!)
Test Ratio:Make a Diagram showing a 'reject' region below

-2.327. Since -0.131 is above this, do not reject
or Critical Value:. Make a Diagram showing a 'reject' region below -.2473. Since -0.0333 is above this, do not reject
or Confidence Interval: is an interval that includes 0. In all cases do not reject .

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The proportions in rows, , are used with column totals to get the items in . Note that row and column sums in are the same as in . (Note that is computed two different ways here - only one way is needed.)

23 23.1 22.9004-0.10000 0.000433

7 6.9 7.1014 0.10000 0.001449

24 23.1 24.9351 0.90000 0.035065

6 6.9 5.2174-0.90000 0.117391

30 30.8 29.2208-0.80000 0.020779

10 9.2 10.86960.80000 0.069565

100 100.0 100.2447 0.00000 0.24468

Since this is less than 9.2103, do not reject .

(Diagram!)

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3. You have a sample of earned incomes for 9 couples, both of whom are teachers. You wish to test if the women make more than the men. (Look at problem 4 before you do this problem - the data are identical, what you do with it is different.)

a. You decide to compare means. Test to see if the women make more than the men assuming that the data comes from a Normal distribution. (5)

b. You are reminded that the income data is usually highly skewed, so you ought to compare medians instead of means. Repeat the test. (5)

For your convenience, not only the raw data are provided, but the data is ranked from 1 to 20 in the columns and . If you need the ranks of the differences, , you will have to do it yourself. Minitab gives us the following results: 61.64, 12.23 (These are the sample mean and sample standard deviation of ), 57.86, 7.56, 3.78, 16.57. You do not need all this information in every part of the problem. Data is in thousands.

Row women rank men rank difference rank

1 47.5945 2 65.6336 15 -18.0391

2 58.5687 8 59.8417 10 -1.2730

3 43.4502 1 59.6314 9 -16.1812

4 62.3263 13 53.1761 6 9.1501

5 60.3484 11 48.1425 3 12.2060

6 65.3845 14 50.6700 4 14.7145

7 80.1389 18 51.5944 5 28.5445

8 78.6558 17 61.0681 12 17.5877

9 58.2903 7 70.9990 16 -12.7087

Solution: This is paired data. a) Assume All tests of the mean or median in Problems three and four are one-sided. The following table may help you choose a method. Save it for the final exam!

Comparing 2 Samples / Paired Samples / Independent Samples
Location - Normal distribution.
Compare means. / Method D4 / Methods D1- D3
Location - Distribution not Normal. Compare medians. / Method D5b / Method D5a
Proportions / Method D6
Variability - Normal distribution. Compare variances. / Method D7

From the outline, there are three ways of approaching a problem involving two means. We know that , 16.57, . We are testing or or , , .

(i) . Confidence Interval: . This interval becomes . Since this interval includes zero, we cannot reject .

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(ii). Test Ratio: . . Make a diagram showing an almost Normal curve with a mean at zero and a 'reject' region above . Since 0.684 is not in this region, we cannot reject .

(iii). Critical Value: . Because this is a one-sided test, we want one critical value above zero. The critical value formula becomes Make a diagram showing an almost Normal curve with a mean at zero and a 'reject' region above 9.734. Since 3.78 is not in this region, we cannot reject .

Note that only one of the three methods above is required.

b) If the underlying distributions are not Normal and the two samples are paired, we should use a Wilcoxon Signed Rank test.

Our hypotheses are . The signed ranks of the differences are at right. If we add ranks with like signs, we get and (Check: Their total is the sum of the numbers 1 through 9, which is ) According to the Wilcoxon table for a 1-sided 5% test, reject the null hypothesis if the smaller of these totals is less than 8.

difference rank

-18.03918-

-1.27301-

-16.18126-

9.15012+

12.20603+

14.71455+

28.54459+

17.58777+

-12.70874-

Since neither of the totals is below 8, do not reject

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4. You have two independent random samples of 9 incomes from each of two towns. You wish to test if the people in town 1 make more than people in town 2. (The numbers just happen to be the same as in the last problem)

a. You decide to compare means. Test to see if the people in town 1 make more than the people in town 2 assuming that the data comes from a Normal distribution. (5)

b. You are reminded that the income data is usually highly skewed, so you ought to compare medians instead of means. Repeat the test. (5)

c. In part a of this question, what assumption did you make about the variances of and ? Test it here. (3).

d. Using the means and variances given above, but assuming that do a 2-sided 98.4% confidence interval for . (4)

Row Town 1 rank Town 2 rank difference rank

1 47.5945 2 65.6336 15 -18.0391

2 58.5687 8 59.8417 10 -1.2730

3 43.4502 1 59.6314 9 -16.1812

4 62.3263 13 53.1761 6 9.1501

5 60.3484 11 48.1425 3 12.2060

6 65.3845 14 50.6700 4 14.7145

7 80.1389 18 51.5944 5 28.5445

8 78.6558 17 61.0681 12 17.5877

9 58.2903 7 70.9990 16 -12.7087

Solution: a) Assume All tests of the mean or median in Problems three and four are one sided. From the outline, there are three ways of approaching a problem involving two means. We know that . We are testing or or . It is most convenient to assume that , though we really ought to test it in part c). It is not wrong to assume that variances differ, but the solution will only be provided to people who did so. If we assume that variances are equal we find , , and .

Only one of the following methods is expected.

(i) . Confidence Interval: . This interval becomes . Since this interval includes zero, we cannot reject .

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(ii). Test Ratio: . . Make a diagram showing an almost Normal curve with a mean at zero and a 'reject' region above . Since 0.7886 is not in this region, we cannot reject .

(iii). Critical Value: . Because this is a one-sided test, we want one critical value above zero. The critical value formula becomes Make a diagram showing an almost Normal curve with a mean at zero and a 'reject' region above 8.368. Since 3.78 is not in this region we cannot reject .

b) If the underlying distributions are not Normal and the two samples are independent, we should use a Mann-Whitney-Wilcoxon Rank test. Our hypotheses are . The rank total for Town 1 (gotten from adding the column) is and, for Town 2, . (Check: Their total is the sum of the numbers 1 through 18, which is ) According to Table 6 (for a 1-sided 5% test), we do not reject null hypothesis if the smaller of these totals () is between 54 and 90. Since both of the totals are in that interval, do not reject

c) . To test this at the 5% significance level, test and against . Since neither is larger than 4.43, do not reject

d) If the confidence level is 98.4%, the significance level is . Since our samples have a total number of degrees of freedom of we can use instead of .

. , and we found that it was 2.41 on page 1. Our confidence interval is