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ECO252 QBA2Name KEY
SECOND HOUR EXAMCircleHour of Class Registered
November 8, 2005MWF2, MWF3, TR12:30, TR3
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.
Make a diagram: For draw a Normal curve with a vertical line at 24 in the middle. Shade the entire area between 0.50 and 30.4. This will cover areas on both sides of 24. Or for draw a Normal curve with a vertical line at zero in the middle. Shade the area from -3.36 to zero and from zero to 0.91.
2.
Make a diagram: For draw a Normal curve with a vertical line at 24 in the middle. Shade the entire area below 34. This will cover areas on both sides of 24. Or for draw a Normal curve with a vertical line at zero in the middle. Shade the entire area below zero and from zero to 1.43.
3.=.4997
Make a diagram: For draw a Normal curve with a vertical line at 24 in the middle. Shade the entire area between zero and 24. This will be an area on the left side of 24. Or for draw a Normal curve with a vertical line at zero in the middle. Shade the entire area from -3.43 to zero.
4.
To findmake a Normal diagram for showing a mean at 0 and 50% above 0, divided into 4.5% above and 45.5% below . So The closest we can come is or . So use (or 1.69 or 1.70). .
Check:
II. (22+ points) Do all the following? (2points each unless noted otherwise). Look them over first – the Computer problem is at the end. Note that some formulas have been squashed by a bug in Word. They should print correctly and will read right if you click on them.
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. Given below is the result from 215 students’ view of the program before and after the orientation.
After the OrientationBefore the Orientation / Good / Not Good / Total
Good / 93 / 37 / 130
Not Good / 71 / 14 / 85
Total / 164 / 51 / 215
- Referring to Exhibit 1, which test should she use?
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
ANSWER:c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: McNemar test, assumption
In Method D6b, the McNemar Test, we compare two proportions taken from the same sample. Assume that two different questions are asked of the same group with the following responses. In this case If we wish to test ,where is the proportion saying ‘yes’ before and is the proportion saying ‘yes’ after, let (The test is valid only if .)
- Referring to Exhibit 1, what is the null hypothesis?
Solution: See above.
- Referring to Exhibit 2, what is the value of the computed test statistic?[6]
Solution: See above. ANSWER:3.37 or 3.37
TYPE: PR DIFFICULTY: Moderate
KEYWORD: McNemar test, test statistic
- Referring to Exhibit 2, what is the p-value of the test statistic using a 5% level of significance? [8]
ANSWER:
0.0008 Actually is less exact, but fine.
TYPE: PR DIFFICULTY: Moderate
KEYWORD: McNemar test, p-value
Exhibit 2 (This was Problem D7)
In a study of sleep gotten with a sleeping pill and with a placebo the results were (Keller, Warren, Bartel, 2nd ed. p. 354)
Pill Placebo difference
7.3 6.8 .5
8.5 7.9 .6
6.4 6.0 .4
9.0 8.4 .6
6.9 6.5 .4
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 .
- Referring to Exhibit 2, what should be the degrees of freedom for this test?
a)
b)*
c)
d)(Rounded to 7.)
e)Degrees of freedom are irrelevant because we should use a (Mann-Whitney-) Wilcoxon rank sum test.
f)Degrees of freedom are irrelevant because we should use a Wilcoxon signed rank test.
g)We do not have enough information to answer this question. (You must explain what information is missing)
- Referring to Exhibit 2, in the formula , we should use the following.
a)
b)
c)*, which is used to compute
d) The standard error is irrelevant because we are using a (Mann-Whitney-) Wilcoxon signed rank test.
e) The standard error is irrelevant because we are using a (Mann-Whitney-) Wilcoxon signed rank test.
f) We do not have enough information to answer this question. (You must explain what information is missing)
- Referring to Exhibit 2, assume that the correct alternate hypothesis is , that use of the
Formula is correct, that and that there are 27
Degrees of freedom (Assume that all of these are correct, even though it is very unlikely!), we should do the following.
a)Reject the null hypothesis only if the significance level is a value below .01
b)Reject the null hypothesis if the significance level is any value below .02
c)Reject the null hypothesis if the significance level is any value above .025
d)*Reject the null hypothesis if the significance is any value above .05
e)Reject the null hypothesis only if the significance level is above .10
f)None of the above.[14]
Explanation: Note and so that, for a 2-sided test, the p-value is between .05 and .02. If the p-value is below the significance level, reject the null hypothesis.
- You are having a part produced in two different machines. Is 201 randomly selected data points that represent the length of parts from machine one, is 501 randomly selected data points that represent the length of parts from machine two. You want to test your suspicion that parts from machine two are more variable in length than parts from machine one (This is the same as saying that machine 1 is more reliable than machine 2). Test this suspicion after stating your hypotheses. Your sample means are 25.593 inches for machine 1 and 25.592 for machine 2. Sample standard deviations are 8.379 for machine 1 and 9.293 for machine 2.(2) [17]
Solution:
Or. In terms of the variance ratio or, the alternate hypothesis rules, so and.
Since you are comparing variances, use Method D7. Compare the ratio against . This has the F distribution with 500 and 200 degreesof freedom. From the table. Since the computed F is larger than the table F, reject the null hypothesis.
- (Extra credit) compute a two-sided confidence interval for the ratio of the two variances in the previous problem. (3)
Solution: From the outline. has
the Fdistribution with 500 and 200 degrees of freedom. So,
. , must be between
and , probably about 1.26. The interval thus becomes
or .
- The following problem is an easier version of a problem in the text.
A pet food canning factory produces 8 oz cans of cat food. The manager suspects that the amount of cat food put into the cans by machine 1 is significantly larger than that put in by machine 2. A sample of output is taken with the results below.
a) What are the manager’s null and alternate hypotheses?(1)
b) You will use a test ratio of the form to test the hypothesis. Find (3)
Warning: Be accurate! is roughly the size of .0000175. If you start rounding excessively, your answers will be completely wrong. If you absolutely cannot do this section, say so and use .000175, which is very wrong.
c) Compute the test ratio and find a p-value for your result. (2)
d) If the manager had, instead, suspected that a larger amount was going into cans filled by machine 2, what would the p-value be? (1)
e) Find a 91% two-sided confidence interval for the difference between the average amount of food put in the cans by the two methods. (2) [25]
Solution: a),so . Note
b)
c)
d) 1-.0183 = .9817.
e) From page 1 (or 1.69 or 1.70), so z
or 0.002 to 0.014.
- Over 104 weeks, the following numbers of mortgages were approved by a bank. Do the results below follow a Poisson distribution?
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Number Approved /0 / 13
1 / 25
2 / 31
3 / 17
4 / 8
5 / 5
6 / 1
7 / 1
8 or more / 0
Total / 104
a) Find the average number of mortgages approved per week? Hint: The original version of the problem came up with a mean of 2.1058 approvals per week. You will come out with something closer to a mean that you actually can find in your tables (1)
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b) If the data followed an appropriate Poisson distribution exactly, what would the numbers of weeks be with 0, 1, 2, 3, 4, 5, 6, 7 and 8 or more approvals be? (3)
c) Use a statistical test to compare the number of actual approvals with the distribution you found in the last section. (3)
d) If all this sounds like too much work, guess the mean and compare the observed data with a Poisson distribution with the mean that you guessed. Do not use a Chi-squared method in d. (5) [37]
Solution: a)
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Number Approved / /0 / 13 / 0
1 / 25 / 25
2 / 31 / 62
3 / 17 / 51
4 / 8 / 32
5 / 5 / 25
6 / 1 / 6
7 / 1 / 7
8 or more / 0 / 0
Total / 104 / 208
This implies
1
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b) The f column to the right was copied from the Poisson table for a parameter of 2. It actually included the following probabilities: for 5, .036089; for 6, .012030; for 7, .003437; for 8, .000859; for 9, .000191; for 10, .000038; for 11, .000007 and for 12.000001. If you multiply these numbers by 104, you will get values of E that are less than 5. They were lumped together in a single class for 5 and over. Note that, if you are rounding these quantities, the easiest way to get the last probability is So
1 0 14.0748 0.135335
2 1 28.1498 0.270671
3 2 28.1498 0.270671
4 3 18.7665 0.180447
5 4 9.3833 0.090224
6 5+ 5.4758 0.052652
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c) The Chi-squared computations are below. Both the traditional and short-cut method are shown.
Row
1 16 14.0748 -1.92516 3.70624 0.263324 18.1885
2 25 28.1498 3.14978 9.92114 0.352441 22.2027
3 31 28.1498 -2.85022 8.12373 0.288589 34.1388
4 17 18.7665 1.76649 3.12048 0.166279 15.3998
5 8 9.3833 1.38330 1.91351 0.203927 6.8206
6 7 5.4758 -1.52419 2.32316 0.424259 8.9485
104 104.0000 0.00000 1.69882 105.6988
The table chi-square for 6 – 1 – 1 = 4 degrees of freedom is . (We lost a degree of freedom because we estimated a parameter from the data.) Our computed chi-square is . Our null hypothesis is that the distribution is Poisson, and since our computed chi-square is less than the table value we cannot reject the null hypothesis.
- A researcher randomly samples female (Sample 1) and male graduates of an MBA program. The figures represent starting salaries. The Minitab command used here is a pull-down command that is equivalent to the two-sample command that you used in your computer assignment.
MTB > TwoT 18 48266.7 13577.63 12 55000 11741.29;
SUBC> Alternative -1.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 18 48267 13578 3200
2 12 55000 11741 3389
Difference = mu (1) - mu (2)
Estimate for difference: -6733.30
95% upper bound for difference: 1229.26
T-Test of difference = 0 (vs <): T-Value = -1.44 P-Value = 0.081 DF = 25
a) Turn in your first computer assignment only. (2)
b) For this assignment make 3 curves with centers at zero and with the t ratio marked by a vertical line. Indicate, using the three diagrams and information from the printout, what the p-value would be if the null hypothesis was
(i)
(ii)
(iii)
Give me a number for the p-value.(1.5)
c) Using the style that I use for null hypotheses, which of the three hypotheses in b) is the null hypothesis used here and would it be rejected if the confidence level was .05? (1)
d) Nothing in the command told Minitab which method to use. Of the four methods you learned to compare two means, which one did Minitab use? (1) [42.5]
Solution:
b)(i)
(ii)
(iii)
Diagram: It says T-Value = -1.44 so make three diagrams with an almost Normal curve and a mean at zero. (i) .919 is the area above -1.44, (ii) .081 is the area below -.144, (iii) .161 is the area above +1.44 and the area below -1.44. (1.5)
c) It says T-Test of difference = 0 (vs <). This says , so the null hypothesis is . P-value is above .05, so do not reject the null hypothesis.
d) Of the four methods you learned to compare two means, Minitab uses D3 when there are independent samples and no explicit directions that say to use a pooled or equal variance method.
ECO252 QBA2
SECOND EXAM
Nov 8-9, 2005
TAKE HOME SECTION
-
Name: ______
Student Number: ______
III. Do sections adding to at least 20 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.) Answers without reasons are not usually acceptable. Neatness counts! Check the website regularly for hints or corrections.
1) A state is trying to figure out whether the background on highway signs makes a difference. In order to do this two samples of 15 individuals are shown a number of slides rapidly. The slides have either a green or a red background. You are trying to find out whether there is a difference between the number of slides correctly read between those with a red or a green background. To do so you will compare the mean or median as appropriate to the distribution. To personalize the data, look at the third digit from the end to decide what red data you will use. Call the column that you pick rj and compute a column called dj with the formula dj = green – rj. (Example: Seymour Butz’s student number is 976512, so he picks column 5 and used d5 = green – r5.) Tell me what column you are using! If you compare means state your hypotheses both in terms of and and in terms of .
Row green r0 r1 r2 r3 r4 r5 r6 r7 r8 r9
1 8 9 5 7 6 5 8 9 7 10 5
2 10 10 9 10 12 6 9 7 9 9 6
3 6 9 5 11 7 5 6 6 10 7 6
4 5 4 7 9 10 6 10 7 11 10 7
5 9 9 6 8 12 5 7 8 11 6 10
6 7 9 7 8 10 11 7 7 6 6 10
7 3 8 9 8 9 8 5 6 11 7 10
8 7 7 8 7 11 8 6 6 6 8 13
9 7 5 7 9 9 7 11 7 9 8 9
10 3 6 5 9 8 9 9 7 7 7 6
11 6 10 7 7 7 10 10 5 7 9 8
12 6 8 5 7 6 7 7 7 7 11 6
13 8 8 8 10 11 7 7 10 9 6 7
14 3 9 7 7 13 6 8 7 8 8 8
15 10 9 7 6 8 5 8 6 8 7 10
Minitab computed some basic statistics from the data which will help you in some parts of this problem.
Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
green 15 6.533 0.601 2.326 3.000 5.000 7.000 8.000 10.000
r0 15 8.000 0.458 1.773 4.000 7.000 9.000 9.000 10.000
r1 15 6.800 0.355 1.373 5.000 5.000 7.000 8.000 9.000
r2 15 8.200 0.368 1.424 6.000 7.000 8.000 9.000 11.000
r3 15 9.267 0.581 2.251 6.000 7.000 9.000 11.000 13.000
r4 15 7.000 0.488 1.890 5.000 5.000 7.000 8.000 11.000
r5 15 7.867 0.435 1.685 5.000 7.000 8.000 9.000 11.000
r6 15 7.000 0.324 1.254 5.000 6.000 7.000 7.000 10.000
r7 15 8.400 0.456 1.765 6.000 7.000 8.000 10.000 11.000
r8 15 7.933 0.408 1.580 6.000 7.000 8.000 9.000 11.000
r9 15 8.067 0.573 2.219 5.000 6.000 8.000 10.000 13.000
a) Display the numbers that you are using in columns and compute a sample mean and sample standard deviation for the d column. (1)
In this problem assume that the red and green data are two independent samples. Use a confidence level of 95%.
b) Assume that you believe that the normal distribution does not apply to the data and compare the means or medians as appropriate. (4)
c) You suspect that the data has the Normal distribution. Test to see if the Normal distribution applies. Use a test that I taught you. (3)
d) You decide that the Normal distribution applies to the data, but do not know if the variances are equal.
Test them for equality. (1)
e) You conclude that the underlying distributions are Normal and that the population variances are equal.
Compare the means or medians as appropriate. Use a test ratio, critical value or a confidence interval (4) or all three (6). [15]
f) (Extra credit) You conclude that the underlying distributions are Normal and that the population variances are not equal. Compare the means or medians as appropriate. Use a test ratio, critical value or a confidence interval (5) or all three (7)
2) In fact the data on the previous page applies to a single sample of 15 individuals. That is the first line of your worksheet tells you how the first person in the sample did when showed the same slides with red or green backgrounds. This applies to a) and b) in this question. Use a confidence level of 95%.
a) Assume that you believe that the normal distribution does not apply to the data and compare the means or medians as appropriate. (3)
b) You assume that the data has the Normal distribution. Compare the means or medians as appropriate. (3)
c) For any part of one of these problems (tell me which one!), compute a confidence interval that you would use to compare means if your alternate hypothesis was (2) [23]
d) For the same part as you used in c), find a p-value for the null hypothesis. (2)[25]
These results are all supposed to look to me as you did them by hand. But what I don’t know won’t hurt me. If you want to check your results by computer, you might try to use the following Minitab routine. If you put green in C1 and label columns with headings like rj, dj and dsqj (Seymour called his green, r5, d5 and dsq5.) The routine below with appropriate changes to rj, dj and dsqj, will compute much of the stuff above, though not in the right order. Note that to do a Wilcoxon signed rank test by hand, you will have to drop all zeroes from the d column.
Computations for comparing c1 and rj
MTB > let dj = c1 – rj
MTB > let dsqj = dj *dj
MTB > print c1 rj dj dsqj
MTB > describe c1 rj dj
MTB > sum dj
MTB > ssq dj
MTB > TwoSample c1 rj;
SUBC> Pooled.
MTB > TwoSample c1 rj.
MTB > Paired c1 'rj'.
MTB > VarTest c1 'rj';
SUBC> Unstacked.
MTB > WTest 0.0 'dj';
SUBC> Alternative 0.
MTB > Mann-Whitney 95.0 c1 'rj';
SUBC> Alternative 0.
If you want to fake the calculations for the Mann-Whitney test, try this.
Procedure for setting up Mann-Whitney Test
#c1 is green, c2 is rj, c3 is difference.
# Mann Whitney Test
MTB > Stack c1 c2 c5;
SUBC> Subscripts c6.
MTB > Rank c5 c7.
MTB > Unstack (c7);
SUBC> Subscripts c6;
SUBC> After;
SUBC> VarNames.
MTB > sum c8
MTB > sum c9
MTB > print c1 c8 c2 c9 #The rest is up to you.
If you want to fake the calculations for the Wilcoxon signed rank test, try this. Unfortunately, I know no good way to remove the zeros or change the signs except by hand.
Procedure for setting up Wilcoxon Signed Rank Test
#c1 is green, c2 is rj, c3 is difference.
MTB > Let c3 = c1-c2#Maybe you already did this.
MTB > # Wilcoxon signed rank test
MTB > let c10 = c3
MTB > #Remove zeroes from c10. (Just use delete on the cells with zeros.)
MTB > #Notice that n has gotten smaller.
MTB > let c11 = abs(c10)
MTB > rank c11 c12
MTB > let c13 = c12
MTB > #Change signs in c13 to agree with signs in c10.
MTB > let c14 = c13 *c10 #Check on signs. All should be positive.