252y0112 2/28/01 ECO252 QBA2 Name ____key______
FIRST HOUR EXAMHour of class registered _____
February 20, 2001Class attended if different ____
Show your work! Make Diagrams!
Note that since relatively few people took this exam, commentary on 252y0111 is more complete.
I. (14 points) Do all the following.
1.
2.
3.
4.
5.
6. A symmetrical interval about the mean with 85% probability. We want
two points , so that. Make a diagram.
If you do a diagram for , it will show two points, and . (which
has 92.5% above it!) is below zero and is above zero. Since zero is halfway
between these two points, the diagram will show 85% split between the two sides
of zero, so that 42.5% is between and zero, and 42.5% is between zero
and . The probability below and the probability above are both 7.5%.
From the diagram, if we replace x by z, . The closest we can come is . So , and or -7.08 to 13.08.
To check this note that
7. We want a point , so that. Make a diagram.
The diagram for will show one point , which has 13.5% above it (and 86.5%
below it!) and is above zero because zero has only 50% below it. Since zero has 50%
above it, the diagram will show 36.5% between zero and .
From the diagram, if we replace x by z, . The closest we can come is . So , and , or 10.70.
To check this note that
252y0112 2/21/01
II. (6 points-2 point penalty for not trying part a.)
The Watched Potts Bank of Pottstown wishes to monitor the debt-to-equity ratio of the firms to which it has made commercial loans. A random sample of 8 of the firms is taken. Assume that the bank is sampling from a population with a normal distribution.
firm Debt/equity ratio
1 1.34
2 1.04
3 1.46
4 1.22
5 1.20
6 1.79
7 1.38
8 1.42
a. Compute the sample standard deviation, , of the debt-to-equity ratios. Show your work! (3)
b. Compute a 95% confidence interval for the mean debt-to-equity ratio, .(3)
Solution: a)
or .
Firm / (Ratio) /1 / 1.34 / 1.7956
2 / 1.04 / 1.0816
3 / 1.46 / 2.1316
4 / 1.22 / 1.4884
5 / 1.20 / 1.4400
6 / 1.79 / 3.2041
7 / 1.38 / 1.9044
8 / 1.42 / 2.0164
Total / 10.85 / 15.0621
b) From the problem statement . From Table 3 of the syllabus supplement, if the population variance is unknown and . . So or 1.170 to 1.542.
252y0112 2/16/01
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) . 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.
Error: On this page, 1.43 should be changed to 1.33. Remove section f!!!
1. The Watched Potts Bank of Pottstown is a subsidiary of the Umongous Bancorp. The Bancorp is threatening to remove the current officers of the bank if the firms to which they are making loans have an average debt-to-equity ratio that exceeds 1.33. Are the officers in trouble? The data are on the previous page. For your convenience the data are repeated below.
firm Debt/equity ratio
1 1.34
2 1.04
3 1.46
4 1.22
5 1.20
6 1.79
7 1.38
8 1.42
Test to see if the mean is above 1.33 using the sample mean and standard deviation you found in part II.
a. Choose the correct null hypothesis from the list below and write the alternative hypothesis. (2)
b. Find a critical value appropriate for this problem, using a confidence level of 95%.(3)
c. Use your critical value to test the hypothesis. State clearly whether you reject the null hypothesis. (2)
d. Repeat the test using (i) a test ratio (2) and (ii) a confidence interval. (2)
e. Test the hypothesis that the average debt-to-equity ratio is exactly 1.33 at the 95% confidence level. (2)
Solution: From Table 3 of the Syllabus Supplement:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueMean (
Known) / / / /
Mean (
Unknown) /
/ / /
a) We are afraid that . Because this does not contain an equality, it cannot be a null hypothesis. So we must test its opposite and our alternate hypothesis is
b) (If you chose a 2-sided test, )
From the previous page: and .
252y0112 2/21/01
Critical Value: Since this is a one-sided test,
or 1.4791. It might help to remember that is always in the 'accept' region.
c) The critical value in b) means that we reject if the sample mean is above 1.4791.
Since is not above this critical value, do not reject .
d) (i) Test Ratio: . This is in the ‘accept’ region
below , so do not reject . It might help to remember
that zero is always in the 'accept' region for
(ii) Confidence Interval: Since this is a one-sided test, and the alternate hypothesis
is ,
or . This does not contradict , because any mean in the
range 1.207 to 1.33 satisfies both statements, so do not reject .
e) , . .
and . Use one of the three methods below.
(i) Critical Value: Since this is a two-sided test,
or 1.1469 and 1.5131. Since is between these two values, do not reject .
(ii) Test Ratio: . This is in the ‘accept’ region
between and , so do not reject .
(iii) Confidence Interval: Since this is a two-sided test, use
. We can thus say that
Since this interval includes we cannot reject .
252y0112 2/21/01
2. Ken Black tells us that, nationwide, 17% of Americans drink milk as their primary breakfast beverage. A milk producer in Wisconsin believes that the proportion of Wisconsin residents that drink milk for breakfast is above 17%. She takes a survey of 500 Wisconsin residents and finds that 105 drink milk as their primary beverage. Use a 90% confidence level in all parts of this problem.
a. Do a 90% confidence interval for the proportion of residents that use milk as their primary beverage. (3)
b. Assume that the milk producer wants to estimate the proportion of Wisconsin residents that use milk as their primary beverage within , how large a sample does she need? Assume that the correct proportion is 17% in this section. (3)
c. Her original purpose in collecting the data presented above was to test the hypothesis that more than 17% of the people in Wisconsin drink milk as their primary breakfast beverage. Test this hypothesis now.
(i) State your null and alternative hypotheses. (2)
(ii) Do a test of your null hypothesis using a test ratio and the 99% confidence level (3)
(iii) Using your test ratio find a p-value for your null hypothesis. (2)
(iv) Do a test of your null hypothesis using a critical value for the observed proportion (2)
(v) Do a test of your null hypothesis using an appropriate confidence interval (2)
d. (extra credit - this was covered in your text, but not in class) How would you modify your work and conclusion in either (ii) or (iii) in c) if you found that your sample of 500 residents had been taken in a state with only 3000 residents? (3)
Solution: From Table 3:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueProportion / / / /
a) and
or .180 to .240.
b) From the outline . and since use . Thus and we must use a sample of at least 15273.
c) (i) Note that , so that .
(ii) Test Ratio: . Since and this is a one-sided test,
use . Since 2.381 is more than , reject . Make a diagram. Show a normal curve with a mean at 0 and a 'reject' region starting at and show where 2.381 falls on this diagram.
(iii) , so reject .
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(iv) Critical Value: . Since is
above this value, reject . Make a diagram. Show a normal curve with a mean at 0.17 and a 'reject' region starting at .1915 and show where .210 falls on this diagram.
(v) Confidence Interval: Since the alternate hypothesis is , the one-sided confidence interval will be . Since contradicts , reject . This is an extremely rare case. Make a diagram. Show a normal curve with a mean at .210 and represent the confidence interval with a region above .1865. Represent the null hypothesis with a region below .17. show that they do not overlap.
d) Because the population size is now , which is less than 20 times the sample size, , we need a finite population correction factor. With this factor, the formula for the variance of the sample proportion becomes .
Note that the text on page 324 omits the -1.
(ii) Test Ratio: . Since and this is a one-sided test,
use . Since 2.607 is more than , reject . Or
(iii) , so reject . The conclusions only became stronger.
252y0112 2/21/01
3. Ken Black says that a soft drink manufacturer takes a sample of 50 12-ounce cans of soda to check that the mean is at least 12 ounces and finds that the sample mean is 11.985 ounces. From long experience, the producer assumes that the population standard deviation is 0.10 ounces. Use a 99% confidence level in a) - d) in this problem.
a. State the null hypothesis and find the p-value for it. (3)
b. Find a critical value for the sample mean. (2)
c. What would the power of the test in b) be if the mean was, in fact, 11.98 ounces? (3)
d. Doing appropriate calculations, find and sketch the power curve for the test in b). (5)
e. Do a 73% confidence interval for the mean based on the sample mean of 11.985 (3)
Solution: a) First, state the problem and find a critical value or values.
and the p-value is .
Note: Since last time you had to use the t table to find the p-value, most of you assumed that you had to do that this time too. Think! Since you were given a population variance, you should have used . Moreover, just because you have to double a p-value in a 2-sided test, doesn't mean you should double it in a 1-sided test.
b) Since this is a one-sided test, the formula for a two-sided critical value becomes , so that . So we will not reject if the sample mean is greater than or equal to 11.9671. Make a diagram. Put 12 in the middle and show a 'reject' region below 11.9671. Shade this area to represent the significance level.
c) Since a type II error is wrongly ‘accepting’ the null hypothesis, we compute the probability that the sample mean will be above or equal to the critical value for each value of . Our computations are below. Note that, in general, for this one-sided hypothesis . Here , so
. The power is 1 - .8186 = .1814.
d) Decide on what values of to use to compute , the probability of a type II error. The usual set of values includes the mean from the null hypothesis, (12), the critical value (11.9671), a point about midway between these values (11.98) and two points, one further out beyond the critical value by a distance about equal to the distance between the null hypothesis mean and the critical value (11.93), and another roughly halfway between this point and the critical value (11.95).
Computefor each value of . Since a type II error is wrongly ‘accepting’ the null hypothesis, we compute the probability that the sample mean will be above or equal to the critical value for each value of . Our computations are below. Note that, in general, for this one-sided hypothesis .
We know that if , we will find and the power will be .01.
At the critical value, and the power is also .5.
252y0112 2/28/01
You now have the power for 5 points below or equal to 12. Make a diagram.
e) From page 3 of this document, , and we know that and . . . To find (You found this on page 1!), make a diagram. Show the mean at zero, that the probability above is .135, so that the probability between zero and must be 50% - 13.5% = 36.5%. So . The best we can do on the Normal table is . So , and or 11.969 to 11.1156.
252y0112 2/21/01
4. (Use a 99% confidence level in all parts of this problem.) My old friend Ken Black talks about a worker who has an average of 20 tubes at his work station. To satisfy the requirements of just-in-time inventory, the superintendent wants the standard deviation of the number of tubes at the station to be less than or equal to 3. The station is visited 8 times and, from the eight measurements taken, she calculates a sample mean number of tubes at the station of 20.875 and a sample standard deviation of 4.5806.
a. For starters, do a 2-sided test of the statement that the population standard deviation is 3. (3)
b. Do a confidence interval for the standard deviation. (3)
c. Actually the test in a) should have been a 1-sided test, so do it again. (2)
d. The superintendent visits the work station 61 times during the month. She now calculates the mean at 20.475 and the sample standard deviation at 3.4520. Repeat the test in a) with these new numbers. (2)
e. Now repeat the test in c) with these new numbers. (2)
f. A paper company will cut farmed timber only if the median height of the trees is over 40 feet. A sample of 40 trees is taken and 27 of them are over 40 feet. State your null and alternate hypotheses and tell whether we would reject the null hypothesis at the 99% confidence level. Finally, tell whether your conclusion means that they do or do not cut. (4)
g. I am sure that the Friendlyville, Alabama police department is writing an average of more than 25 speeding tickets a day on its one-block-long main street. Sho' 'nuff , in a one-day visit I see 38 tickets written. Assuming that the Poisson distribution applies, state the null and alternative hypotheses and test the null hypothesis. (3)
h. (Extra Credit) We agree that one day is not enough time to test the hypotheses in g) and so I stay for a full week. In that time I see 185 tickets written. Repeat the test. (3)
Solution: a) or and
From the outline, the test ratio formula is . From the chi-square table, and . Make a diagram. Show a chi-squared distribution with a mean at about , and a 99% 'accept region between the two values of chi-squared just given. Show one 2.5% 'reject' region below the 99.5% value and a second 'reject' region above the 0.5% value. Since 16.31925 falls in the 'accept' region, do not reject .
b) According to the outline, a two-sided confidence interval would be . If we fill in the blanks, we get or . If we take square roots, we get .
c) or and Make a diagram. Show a chi-squared distribution with a mean at about , and a 99% 'accept region below . Show the 1% 'reject' region above the 1% value of chi-squared. Since = 16.31925 falls in the 'accept' region, do not reject .
252y0112 2/21/01
d) or and . For large samples . Make a diagram. Show a normal curve with a mean at 0 and 'reject' regions below and above and show where 1.6920 falls on this diagram. Do not reject .
e) or . Make a diagram. Show a normal curve with a mean at 0 and a 'reject' region above and show where 1.6920 falls on this diagram. Do not reject .
f) From outline point B6a for the sign test for a median:
Hypothesis aboutHypotheses about a proportion
a medianIf is the proportionIf is the proportion
above below
We are told that are above 40 feet and that . So becomes and there are several methods to do this in B6 and the problem solutions. If we use the formula used by the text (quoted as in the outline) without a continuity correction we find . Since this is below we reject and cut the timber.
g) From the Poisson(25) table, . Since this is below we reject .
h) This time we follow Problem B4d) and say (using the Normal approximation to the Poisson distribution) . Since this is not below we do not reject .
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