2/25/00 252y0012 ECO252 QBA2 Name ______KEY______
FIRST HOUR EXAM Hour of class registered __
February 22, 2000 Class attended if different _
Show your work! Make Diagrams! Note that all diagrams are available on a separate sheet.
I. (14 points) Do all the following.
1.
2.
3.
4.
5.
6. A symmetrical interval about the mean with 83% probability. We want
two points , so that. From the diagram,
if we replace x by z, . The closest we can come is
. So , and ,
or -7.59 to 11.59. To check this note that
7. We want a point , so that. From the diagram,
if we replace x by z, . The closest we can come is
. So , and ,
or 9.91 . To check this note that
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II. (6 points-2 point penalty for not trying part a.)
A truck dealer asserts that a new model of truck has lower gas consumption than your present fleet model. A random sample of 7 runs is made with the new model with gas consumption as below. Assume that we were sampling from a normal distribution.
run Consumption
1 11.56
2 10.97
3 12.74
4 12.08
5 12.66
6 12.84
7 11.15
a. Compute the sample standard deviation, , of gasoline consumption. Show your work! (3)
b. Compute a 90% confidence interval for the mean gas consumption, .(3)
Solution: a.
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or .
Home / (Days) /1 / 11.56 / 133.6336
2 / 10.97 / 120.3409
3 / 12.74 / 162.3076
4 / 12.08 / 145.9264
5 / 12.66 / 160.2756
6 / 12.84 / 164.8656
7 / 11.15 / 124.3225
Total / 84.00 / 1011.6722
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.
b. From the problem statement . From Table 3 of the syllabus supplement, if the population variance is unknown and . . So or 11.425 to 12.575.
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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) . Show your work! State and where appropriate. Use a 95% confidence level unless another level is specified.
1. A truck dealer asserts that a new model of truck has lower gas consumption than your present fleet model. The data is on the previous page. On the basis of a great deal of experience, you know that the mean for your present model is 12.3 miles per gallon. For your convenience the data are repeated below.
run Consumption
1 11.56
2 10.97
3 12.74
4 12.08
5 12.66
6 12.84
7 11.15
Evaluate the dealer’s statement 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 90%.(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 mean consumption of gas is exactly 12.1 for the dealer's fleet at the 90% confidence level. (2)
f. Assume that the data does not come from a normal distribution and evaluate the statement that the median is 12.8. (4)
Solution: From Table 3 of the Syllabus Supplement:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueMean (s
Known) / / / /
Mean (s
Unknown) /
/ / /
a. The dealer's statement is . 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.
From the previous page: and .
Critical Value: Since this is a one-sided test,
or 11.87. It might help to remember that is always in the 'accept' region.
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c. The critical value in b) means that we reject if the sample mean is below
11.87. Since is not below this critical value, do not reject .
d. (i) Test Ratio: . This is in the ‘accept’ region
above , so do not reject . It might help to remember
that zero is always on 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 12.3 to 12.4258 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 11.525 and 12.675. 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 .
f. , . If we subtract 12.8 from the values of x, we get the following:
run
1 11.56 -1.24
2 10.97 -1.83
3 12.74 -0.06
4 12.08 -0.72
5 12.66 -0.14
6 12.84 +0.04
7 11.15 -1.65
For , there is 1 value above the alleged median and 6 values below. The p-value is thus
either or . From the binomial table for Since this is above the significance level we cannot reject .
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2. Using data from part II
a. Do a 90% confidence interval for the standard deviation for your fleet (3)
b. Test the hypothesis that the population standard deviation is less than 1 mile per gallon at the 90% confidence level. (4)
c. Using the sample standard deviation that you got in part II, test the hypothesis that the population standard deviation is 1.2 at the 90% confidence level if the sample size is 62. (3)
d. Assuming that the population standard deviation is 0.9 miles per gallon and using the sample mean that you found in part II, test again the null hypothesis about the mean that you found in question III-1a on the previous page. (2)
e. Repeat 2d above assuming that the data come from a sample of 7 out of a population of 70. (2)
Solution: From part II , or . Also
and
a. From page 1 of the Syllabus Supplement: . Since and , look up and . So or or . Finally, taking square roots, .
b. . From the outline is the test ratio. If
and is true, will tend to be below 1, and the ratio will be on the small side.
Since this is a 1-sided test and ,compare the test ratio with
Since the mean of this chi-squared distribution is and the mean must be in the 'accept'
region, our test is to reject the null hypothesis if We compute so we cannot reject .
c. . or , but From the outline is the test ratio. Since this is a large sample, . , - we reject the null hypothesis if is not between Since -3.799
is below that range, reject You could also use a confidence interval here.
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d. , . ,
and . Use one of the three methods below.
(i) Critical Value: Since this is a one-sided test,
= 11.88. Since is above the critical value, do not reject .
(ii) Test Ratio: . This is not in the ‘reject’ region
below , so do not reject .
(iii) Confidence Interval: Since this is a one-sided test, use
. But does not contradict
, so we cannot reject .
e. , . ,
and . Use one of the three methods below.
(i) Critical Value: Since this is a one-sided test,
= 11.883. Since is above the critical value, do not reject .
(ii) Test Ratio: . This is not in the ‘reject’ region
below , so do not reject .
(iii) Confidence Interval: Since this is a one-sided test, use
. But does not contradict
, so we cannot reject .
Remember! My most common comment in grading was "Did you read the question?" If you answer a question about the standard deviation with an answer appropriate for a proportion or a mean, no amount of knowledge of the formulas will help you!
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3. We have the following data (Assume in this question that the original data comes from the normal distribution.):
and .
a. What is the smallest sample size that we could use so that a 90% confidence interval for the mean could
be given with an error of ? (3)
b. If we test the hypothesis that the population mean is 975, what is the p-value for this hypothesis? (3)
c. If our confidence level is 94%, would we reject the hypothesis in 3b? Why? (1)
d. If instead your data reads and , find a range for the p-value and tell
whether you would reject the null hypothesis if the confidence level is 94%. (3)
e. Due to the advent of EasyPass the mean number of cars passing a Garden State Parkway toll booth with
a human collector is falling. Assume that the average number of motorists passing a given location was
known to be 10.5 per minute. Do a one sided hypothesis test using a confidence level of 95% if an
observer reports that the number that actually passed in a minute was 5. (3)
f. To get a more powerful test note that 10.5 cars a minute means 630 per hour. Would you reject the null
hypothesis if 300 cars passed in an hour? (3)
Solution: a. and . From the outline . So use a sample size that is at least 1083.
b. , . , and .
c. If We reject a hypothesis if the p-value is below the significance level.
Since this is not true here, do not reject .(To do it the hard way use .)
d. . , . and , , From the t table = 2.745 is smaller than ,
but is larger than , so that, for a 1-sided test, we can say .
But for this 2-sided test we must say . We reject if
Since we reject .
e. Since this is a problem involving a comparison of the number per unit time to the average per unit time we use the Poisson distribution. , . This is a 1-sided test, so look up on the table. You should get a p-value of .05038. Since this is above the significance level of .05, do not reject the null hypothesis.
f. Since this is a problem involving a comparison of the number per unit time to the average per unit time we use the Poisson distribution. , . This is a 1-sided test, so we would like to look up on the table. Because we do not have this table we approximate the Poisson distribution with a Normal distribution with and . Thus . Since this is below the significance level of .05, reject the null hypothesis
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4. An electronics supplier claims that at least 96% of parts supplied to a government agency meet specifications. A sample of 400 parts is tested and 36 parts are rejected.
a. State the null and alternative hypotheses (2).
b. Test the hypothesis at the 95% level. (3)
c. If we wish to create a 2-sided confidence interval for the proportion that meet specifications with an
error of not more than , what is the minimum sample size required? (3)
d. To check the correctness of your solution in 4c, create the 2-sided confidence interval. (If you did not
do 4c, assume a sample size of 237.) (2)
e. Remembering what you did on page 1, do a 74% confidence interval for the proportion. (2)
Solution: From Table 3:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueProportion / / / /
a. Note that , so that , or .
b. (i) Test Ratio: . Since and this is a one-sided test,
use . Since –5.103 is less than , reject . Or
, so reject . Or:
(ii) Critical Value: . Since is
below this value, reject . Or:
(iii) Confidence Interval: . Since
contradicts , reject .
c. From the outline . and since , use Thus and we must use a sample of at least 3147.
d. and .
e. Note that so that and, from the front page, or .904 to .916.
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