10/28/03 252y0323 ECO252 QBA2 Name KEY
(Page layout view!) SECOND HOUR EXAM Hour of Class Registered
October 30, 2003 Circle 11am 12:30pm
I. (53 points) Do all the following (2points each unless noted otherwise).
(Note that some equations have been ‘squashed’ by a bug in Minitab. They print out just fine, but you may have to click on them to see them on your screen.)
- A manufacturer revises a manufacturing process and finds a fall in the defect rate of
a) *The fall in defects is statistically significant because 5% is larger than 4%.
b) The fall in defects is statistically significant because the confidence interval supports H0.
c) The fall in defects is not statistically significant because 4% is smaller than 5%.
d) The fall in defects is not statistically significant because the confidence interval would lead us to reject H0.
Explanation: as a confidence interval means the interval .01 to .09. Since this does not include zero the values are significance. Formally, we are testing with a confidence interval. If we reject the null hypothesis, the difference is significant.
2. If we wish to determine whether there is evidence that the proportion of successes is higher in group 1 than in group 2, the appropriate test to use is
a) *the test. is always a test using .
b) the test.
c) both of the above
d) none of the above
TABLE 12-14
Recent studies have found that American children are more obese than in the past. The amount of time children spend watching television has received much of the blame. A survey of 100 ten-year-olds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance.
Weights / TV Hours / Total0-3 / 3-6 / 6+
More than 10 lbs. overweight / 1 / 9 / 20 / 30
Within 10 lbs. of normal weight / 20 / 15 / 15 / 50
More than 10 lbs. underweight / 10 / 5 / 5 / 20
Total / 31 / 29 / 40 / 100
3. Referring to Table 12-14, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending 3-6 hours, on average, watching TV and are more than 10 lbs. underweight?
a) 5
b) *5.8
c) 6.2
d) 8
Explanation: In the total column 20 out of 100 are more than 10 lbs underweight. 20% of 29 is
5.8.
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4. Turn in your computer output from computer problem 1 only tucked inside this exam paper. (3 points - 2 point penalty for not handing this in.)
MTB > TwoT 90.0 'educ' 'sex';
SUBC> Alternative -1.
Two-Sample T-Test and CI: EDUC, SEX
Two-sample T for EDUC
SEX N Mean StDev SE Mean
Female 788 13.19 3.03 0.11
Male 651 13.28 2.85 0.11
Difference = mu (Female) - mu (Male )
Estimate for difference: -0.091
90% upper bound for difference: 0.108
T-Test of difference = 0 (vs <): T-Value = -0.58 P-Value = 0.280 DF = 1412
The computer output above refers to a test very much like the Minitab test you ran of two independent samples. The major difference is that 1439 numbers appear in column 1 (labeled EDUC) which give number of years of education completed and the computer sorted them by gender using the words ‘female’ and ‘male’ in column 5 (labeled SEX). The variable can thus refer to an imaginary column of female education figures and to in imaginary column of male education figures. Call this the GSSEduc output.
5. Referring to the GSSEduc output, and using the rules taught in class, the null hypothesis that was tested is .
a) H0:
b) *H0:
c) H0:
d) H0:
Explanation: On the last line of the output it says “T-Test of difference = 0 (vs <).” So we have (Since can’t be a strict inequality). On the 4th line from the bottom, it says “Difference = mu (Female) - mu (Male )”. This says If we put these together we get The opposite is
6. Referring to the GSSEduc output, we can conclude, (doing no more calculations) that, for the particular population that was sampled
a) At the level, there is sufficient evidence that women had fewer years of education than men.
b) At the level, there is a difference between the years of education gotten by men and women.
c) *At the there is insufficient evidence that the average men’s education level is higher than the women’s.
d) At the level, there is sufficient evidence to conclude that there is no difference between men’s and women’s education level.
Explanation: We cannot reject the null hypothesis because the p-value is above the significance level. Saying ‘we cannot reject’ is equivalent to saying ‘insufficient evidence to reject.’
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7. Referring to the GSSEduc output, the most commonly used methods to find degrees of freedom are (i) to calculate , or (ii) to say that since we have large sample to use , which is equivalent to saying that the degrees of freedom are infinite, yet the computer claims . Explain, briefly, what the computer probably did (and assumed) to get that number.
Solution: From the syllabus supplement.
Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical ValueDifference
Between Two
Means(s
Unknown,
Variances
Assumed
Unequal) /
/ / /
The computer assumed that it was comparing two independent samples with (possibly) unequal variances (Method D3). It used the formula shown as above.
8. (Wonnacott and Wonnacott) A small piece of hose in the cooling system of a new engine has a lifetime that varies normally (following the Normal distribution) around a mean of 18 months with a standard deviation of 4 months. The first maintenance check occurs at 12 months. What is the probability that the hose will wear out before the maintenance check? (This is the same as the per cent of hoses that will wear out before the first maintenance check!) Make a diagram!
Solution: .
Make a diagram! Either a diagram for with zero in the middle and the area below -1.5 shaded or a diagram for with 18 in the middle and the area below 12 shaded.
9. In problem 5 above, the manufacturer decides that too much money is being spent on maintenance checks. If the manufacturer is willing to accept having 20% of hoses wear out before the fist maintenance check, how many months (to the nearest 100th of a month) can the manufacturer wait until the check? (This is the same as finding the 20th percentile of the distribution)
Solution: . We want , the 20th percentile of this distribution. The easy way is to note that according to the t table . This means that . So it must be true that , or Using the formula , we get . The hard way to do this, which we can only avoid because .20 is an easy number to find on the t table, is to make a diagram for with zero in the middle and an area marked 50% above zero . We know is below zero since 20% is below , and 50% is below zero. There must be 30% between and zero. But we know , so If we look at the Normal table, the closest we can come is This implies that and that .
Check: Make a diagram!
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10. The t test for the difference between the means of 2 independent populations assumes that the respective
a) sample sizes are equal.
b) sample variances are equal.
c) *populations are approximately normal.
d) all of the above
TABLE 10-3
The use of preservatives by food processors has become a controversial issue. Suppose 2 preservatives are extensively tested and determined safe for use in meats. A processor wants to compare the preservatives for their effects on retarding spoilage. Suppose 15 cuts of fresh meat are treated with preservative A and 15 are treated with preservative B, and the number of hours until spoilage begins is recorded for each of the 30 cuts of meat. The results are summarized in the table below.
Preservative A Preservative B
= 106.4 hours = 96.54 hours
= 10.3 hours = 13.4 hours
11. Referring to Table 10-3, state the test statistic for determining if the population variance for preservative B is larger than the population variance for preservative A.
a) F = 3.100
b) F = 1.300
c) *F = 1.693
d) F = 0.591
Explanation: We have the alternative hypothesis . Since the rule says to put
the larger variance in the alternate hypothesis on top, we get .
12. Referring to Table 10-3, what assumptions are necessary for a comparison of the population variances to be valid?
a) Both sampled populations are normally distributed.
b) Both samples are random and independent.
c) Neither (a) nor (b) is necessary.
d) *Both (a) and (b) are necessary.
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TABLE 10-4
A real estate company is interested in testing whether, on average, families in Gotham have been living in their current homes for less time than the families in Metropolis have. A random sample of 100 families from Gotham and a random sample of 150 families in Metropolis yield the following data on length of residence in current homes:
Gotham: = 35 months, = 900 Metropolis: = 50 months, = 1050
13. Referring to Table 10-4, which of the following represents the relevant hypotheses tested by the real estate company?
a) *
b)
c)
d)
Explanation: The problem statement starts out by saying we want to test . This is an alternative hypothesis because it is a strict inequality. If is below , will be negative.
14. Referring to Table 10-4, what is the estimated standard error of the difference between the two sample means?
a) *4.00
b) 4.06
c) 5.61
d) 8.01
e) 16.00
Explanation: These are humongous samples so we use method D1. For this method we have
15. Referring to Table 10-4, what is (are) the critical value(s) for the test ratio for the relevant hypothesis test if the level of significance is 0.05?
a) * = – 1.645
b) = 1.960
c) = – 1.960
d) = – 2.080
Explanation: Since the alternative hypothesis is , this is a left-tail test and, if we use a test ratio we will compare it with
16. When testing versus , the observed value of the -score (test ratio) was found to be – 2.13. The p-value for this test would be
a) 0.0166.
b) 0.0332.
c) 0.9668.
d) *0.9834.
Explanation: Since the alternative hypothesis is , this is a right-tail
test and, if we use a test ratio we will compute . Since this is a right-tail
test, we need
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TABLE 10-9
A buyer for a manufacturing plant suspects that his primary supplier of raw materials is overcharging. In order to determine if his suspicion is correct, he contacts a second supplier and asks for prices on various materials. He wants to compare these prices with those of his primary supplier. The data collected is presented in the table below, with some summary statistics presented (all of these might not be necessary to answer the questions which follow). The buyer believes that the differences are normally distributed and will use this sample to perform an appropriate test at a level of significance of 0.01.
Primary Secondary
Material Supplier Supplier Difference
1 $55 $45 $10
2 $48 $47 $1
3 $31 $32 – $1
4 $83 $77 $6
5 $37 $37 $0
6 $55 $54 $1
Sum: $309 $292 $17
Sum of Squares: $17,573 $15,472 $139
17. Referring to Table 10-9, the hypotheses that the buyer should test are a null hypothesis that (Fill in blanks) or or versus an alternative hypothesis that or or . The text says vs. .
18. Referring to Table 10-9, the test to perform is a
a) pooled-variance t test for differences in 2 means (D2).
b) separate-variance t test for differences in 2 means (D3).
c) Wilcoxon signed rank test for differences in 2 medians (D5b).
d) *t test for mean difference in paired data (D4).
e) Wilcoxon-Mann-Whitney test for differences in 2 medians (D5a).
19. Referring to Table 10-9, the number of degrees of freedom is
a) *5.
b) 10.
c) Irrelevant because you are using a rank test.
d) Found by a complicated formula
Explanation: There are 6 pairs of numbers.
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20. Two brands of gasket are being considered are for use on a high pressure oil pump. The number of hours that the gasket worked are as follows.
Pump Brand 1 Brand 2 difference
1 2982.28 2863.39 118.89
2 3025.86 2906.97 118.89
3 2952.02 2873.52 78.50
4 2954.64 2959.06 -4.42
5 2981.01 2899.98 81.03
Because the data is paired, a test was run using Minitab,(method D4) with the following results
MTB > Paired c7 c8;
SUBC> Alternative 1.
Paired T-Test and CI: brand 1, brand 2
Paired T for brand 1 - brand 2
N Mean StDev SE Mean
brand 1 5 2979.2 29.7 13.3
brand 2 5 2900.6 37.3 16.7
Difference 5 78.6 ______
95% lower bound for mean difference: 30.6
T-Test of mean difference = 0 (vs > 0): T-Value = 3.49 P-Value = ?
Compute the standard deviation of the column, showing your work, and fill in the blanks in the difference row. You should get a t-ratio approximately equal to the T-Value shown above. State the hypotheses, find an approximate p-value and tell whether you reject the null hypothesis. (7)