3/22/00 252y0021ECO252 QBA2Name _____key______
SECOND HOUR EXAMHour of class registered _____
March 21,2000 Class attended if different ____
Show your work! Make Diagrams!
I. (14 points) Do all the following.
1.
2.
3.
4. (The cumulative probability up to 15)
5.
6. A symmetrical interval about the mean with 48% probability.
We want two points , so that. From the diagram,
if we replace x by z, . The closest we can come is
or . Since these are about the same
distance from .2400 use , and ,
or 5.13 to 12.87. 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 17.22 .
To check this note that
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II. (6 points - 2 point penalty for not trying)
A manufacturer of mowers wants to compare the amount of time it takes to mount a mower engine using two different processes. Two independent samples are taken and the time in minutes appears below. ! Note: Most of you didn't bother to tell me what your hypotheses were in any problem but the first one. This means that I couldn't really tell whether your critical values etc. were right or wrong!
2.13.3
4.17.3
9.15.3
3.18.3
2.14.3
3.3
Note that and
a. Compute the sample standard deviation for the second process, . Show your work.(3)
b. Compute a 95% 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)
a. Solution: and
1
3.3 10.89
7.3 53.29
5.3 28.09
8.3 68.89
4.3 18.49
3.3 10.89
31.8 190.54
. .
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b. From page 10 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 -4.62 to 2.22. The
interval includes 0, so there is no significant difference between the means.
Formally, our hypotheses are or or We do not reject.
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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. For your convenience, the data from the previous page, giving samples of the time to mount a mower engine, is copied below. Test the hypothesis that the mean time to mount a mower is greater for method two than method one.
2.13.3
4.17.3
9.15.3
3.18.3
2.14.3
3.3
Note that and
a) Choose the correct null hypothesis from the list below and write the alternative hypothesis. (2)
b) Assume that and 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)
find an approximate p-value for your null hypothesis. (1)
e) Do a 90% confidence interval for the ratio of the standard deviations of the populations from which and come. On the basis of this comparison, what assumption should we have made about the equality of the variances? (3)
Solution: a) The hypothesis says that . Since this does not include an equality, it is the alternate hypothesis. The null hypothesis is the opposite, , so that our hypotheses are or or where . Did you make the same mistake on this exam as you did in the first one?
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b) and , and .
We have already computed = .
Critical Value: is the formula for a 2-sided critical value, but
since our alternative hypothesis is or , we want a
critical value below zero and to use So
c) is above this value, so we do not reject. Unless you gave me a
clear indication of what the rejection region was, e.g. a diagram, and unless
the reject region did not include you did not get full credit on c) or d) .
d) (i) Test Ratio:. Since this is above ,
we do not reject.
(ii) Confidence Interval: The 2-sided interval is , but since our alternative
hypothesis is , we use
Since does not contradict , we do not reject.
(iii) . This value of is between and
so we can say that
e) A 2-sided confidence interval is or , and and . So and . which means that and . If we use the first formula, reduces to and, if we take square roots, . If we use the second formula, , and . Since either of these intervals includes 1, we can say that the variances are not significantly different.
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2. The New York Times reported on exit polls taken of voters in 6 Republican primaries. The numbers and per cent that voted for George W. Bush in four of these are shown below. (Use a 99% confidence level.)
State / Massachusetts / California / Connecticut / New YorkSample Size / 485 / 1361 / 968 / 1060
Number for Bush / 262 / 857 / 542 / 604
Proportion / .540 / .630 / .560 / .570
a)Test the hypothesis that the proportion for Bush in New York was higher than the proportion in Connecticut. (4)
b) Test the hypothesis that the proportion for Bush was the same in each state. (7)
Solution: a) From Table 3 of the Syllabus Supplement: Let be the proportion in CT and be the proportion in NY.
/ Test Ratio / Critical Value
Difference
between
proportions
/
/
/
Or use /
Same as or ,
, ,
(Only one of the following methods is needed!)
Test Ratio: This is above -2.327. (Diagram!)
or Critical Value:
is above this value.
or Confidence Interval: where . So . The interval includes 0. In all cases do not reject .
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b)
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.)
262 283.58 242.062-21.5800 1.64220
223 201.42 246.892 21.5800 2.31207
857 795.78 922.930 61.2200 4.70970
504 565.22 449.411-61.2200 6.63084
542 565.99 519.027-23.9900 1.01684
426 402.01 451.422 23.9900 1.43161
604 619.78 588.622-15.7800 0.40177
456 440.22 472.34615.7800 0.56565
38743874.003892.712 0.000018.711
Since this is less than 11.3449, reject .
(Diagram!)
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3. We have the following data:
a) If we test the hypothesis that the population mean is 978, what is the p-value for this hypothesis? (2)
b) If we use a significance value of 10% and the p-value in a), would we reject the null hypothesis? Why?
(1)
c) Find the critical values of necessary to test the hypothesis in a) if the significance level is 10%. (2)
d) Do a power curve for the test in c). (6)
e) A table of solid-waste generation rates for countries classified by level of development was assembled. We are hypothesizing that as countries' incomes rise they become more alike in their tendency to generate solid waste. For 5 industrial countries the standard deviation for a measure of solid waste generation was 0.0652 and, for 7 middle-income countries, the standard deviation was 0.1828. State an appropriate hypothesis and test it. (4)
Solution: a) From Table 3 of the Syllabus Supplement:
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueMean (
Known) / / / /
b) If , so reject .
c) or 971.42 to 984.58.
d) The table below is the same as used in Problem C2 and summarizes our results and further calculations. means a lower critical value and means an upper critical value. These become and . Note that when power, and when Power . These do not have to be computed.
/ / / / / / Power978 / / / -1.645 / 1.645 / .4500 + .4500 = .9000 / 10.00%
981.3 (Same as 974.7) / / / -2.47 / 0.82 / .4932 + .2939 = .7871 / 21.29%
984.58 (Same as 971.42) / / / -3.29 / 0.00 / .4995 / 50.00%
987.9 (Same as 968.1) / / / -4.12 / -0.83 / / 79.67%
991.2 (Same as 964.8) / / / -4.945 / -1.655 / / 94.00%
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Diagram:
e) . and . Since , , and , if you used a significance level of .10 or .05 do not reject . But if you used a significance level of .025 or .01, reject .
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4. A hospital administrator wishes to compare the distribution of unoccupied beds in two hospitals. Because she believes that the distributions are quite badly skewed she does not use a method based on means but instead ranks the entire sample and uses a test based on these ranks. Data is below. (Source W-890)
Hosp 1Beds / rank / Hosp 2
Beds / rank / Difference
Day / / / / /
1 / 6 / 3 / 34 / 16 / -28
2 / 38 / 17 / 28 / 12 / 10
3 / 3 / 1 / 42 / 20 / -39
4 / 17 / 9 / 13 / 6 / 4
5 / 11 / 5 / 40 / 19 / -29
6 / 30 / 13 / 31 / 14 / -1
7 / 15 / 7 / 5 / 2 / 10
8 / 16 / 8 / 32 / 15 / -16
9 / 25 / 10 / 39 / 18 / -14
10 / 9 / 4 / 27 / 11 / -18
Sum / 77 / 133
Note: The two rank sums were not in the original problem.
a. On the basis of the rank sums (and assuming that the data represents two independent samples,) test the hypothesis that the median number of beds unoccupied differs for the two hospitals. (5)
b. A statistician claims that her method is inappropriate because she has ignored the fact that each row corresponds to a specific day, even if the days were chosen randomly. Rerank the data to account for the fact that it is cross-classified and repeat the analysis. (5)
c. Assume instead that this data is normally distributed paired data, test the hypothesis that the mean number of beds unoccupied differs. You may need some of the following data:
and (4)
Solution: a) Wilcoxon-Mann-Whitney Method . If we sum the two rankings, we get and . Check: . For a 5% two-tailed test with , Table 6 says that the lower critical value is 79. The lower of the two rank sums, is less than this value, so reject
b. Wilcoxon Signed rank test for paired data. .
1
difference rank
-28 8 -
10 3.5+
-3910 -
4 2 +
-29 9 -
-1 1 -
10 3.5+
-16 6 -
-14 5 -
-18 7 -
If we add items with + and – signs separately, we find . To check this, compute . From Table 7 with , , and since 9, the smaller is above the critical value, do not reject
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- Test of equality of means for paired data.
or or
Test Ratio: This is on the interval
between –2.262 and +2.262.
or Critical Value:
is on this interval.
or Confidence Interval: or –24.43
to 0.23. This interval includes 0.
With all methods do not reject . Note that this method is more powerful than the one in c. However, it still should not be used unless the conditions justify it.
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Extra credit. Repeat the following sections of problem III 1) assuming that . . Test the hypothesis that the mean time to mount a mower is greater for method two than method one.
b) Find a critical value appropriate for this problem, using a confidence level of 90%(4)
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)
Solution: b) From the formula table;
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueDifference
between Two
Means(
unknown,
variances
assumed
unequal) /
/
Same as
/ /
and , and . . or or where .
, so use 7 degrees of freedom.
Critical Value: is the formula for a 2-sided critical value, but since our alternative
hypothesis is , we want a critical value below zero.
So
c) is above this value, so we do not reject.
d) (i) Test Ratio:. Since this is above ,
we do not reject.
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(ii) Confidence Interval: The 2-sided interval is , but since our alternative
hypothesis is , we use
Since does not contradict , we do not reject.
1