IE241 Final Exam

1. What is a test of a statistical hypothesis?

Decision rule to either reject or not reject the null hypothesis.

2. In a poll taken among college students, 50 of 200 women favored a certain proposition, whereas 100 of 300 men favored it. Is there a real difference of opinion on this proposition?

Ho: pf = pm

Ha: pf ≠ pm

Critical value of t = 1.96 for this 2-tailed test at α = .05 because t = z because of large sample sizes

50/200 = 0.25 female proportion 100/300 = 0.33 male proportion 150/500 = 0.30 overall proportion

Because the observed t > -1.96, we cannot reject the null hypothesis.

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3. What are the components of the decision rule?

test statistic and critical region

4. Two sets of 12 children each were taught speed swimming by two different methods. After the instruction was over, each class swam ten laps in the pool. For group A, the mean time = 8 minutes and the variance = 9. For group B, the mean = 6 minutes and the variance = 16. Is there a difference between the two teaching methods?

Note: Hypotheses are always stated in Ho: μA = μB terms of parameters, so you can’t have

Ha: μA ≠ μB Ho:

The critical value of t =2.074 for this 2-tailed test with (12-1)+(12-1) df at α = .05

Because the observed t < 2.074, we cannot reject the null hypothesis.

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5. What is the logic behind the use of the critical region?

Under the null hypothesis, the test statistic can fall anywhere in its distribution. But the probability α that the test statistic will fall in the critical region (one of its extreme tails) is very small so if it does fall there, we choose not to believe in the null hypothesis.

Note: Many of you answered this by describing the use of the critical region (rejection of Ho), instead of the logic behind its use. The question doesn’t ask how it’s used, but why it is used that way.

6. Of 64 offspring of a certain cross between guinea pigs, 34 were red, 10 were black, and 20 were white. According to the genetic model, these colors should be in the ratio 9:3:4. Are the data consistent with the genetic model?

Ho: data are consistent with model 9:3:4 Ha: data are not consistent with model 9:3:4

For this goodness-of-fit test with (3-1) df at α=.05, the critical value of χ2 = 5.99. Since the observed value of χ2 < 5.99, we cannot reject Ho. These data do not support the conclusion that the data are inconsistent with the model.

red black white total

Obs freq 34 10 20 64Exp freq

(9/16)64 =36

(3/16)64 =12

(4/16)64 =16

64

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7. Explain the two types of error when we do a hypothesis test.

If we reject the null hypothesis when it is fact true, we make a type 1 error whose probability α we can control.

If we do not reject the null hypothesis when it is in fact false, we make a type 2 error, whose probability β is minimized for all test with the same α.

8. Given that X is normally distributed, assume that the results of a sample of size 16 had a mean = 62 and s = 10. (a) Test the hypothesis that σ = 8. (b) Test the hypothesis that μ = 50.

(a) Ho: σ = 8 Ha: σ ≠ 8 Note: This is a 2-tailed test because it asks “is σ = 8 or not.” Critical value of χ2 = 25.0 with (16-1) df at α= .05. Since the observe

d χ2 < 25.0, we cannot reject Ho.

(b) Ho: μ = 50 Ha: μ ≠ 50 Note: This is a 2-tailed test because it asks “is μ = 50 or not.”

Critical value of t = 2.131 for this 2-tailed test at α = .05 with 15 df. Since the observed value of t > 2.131, we reject the null hypothesis and declare that the mean ≠ 50.

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9. Why can we not accept the null hypothesis if the test statistic does not fall in the critical region?

We never accept the null hypothesis, but only fail to reject it with the data at hand because a different, more powerful test may yield data which will enable us to reject the null hypothesis.

10. A cigarette manufacturer sent each of two laboratories samples of tobacco. Each lab made 5 determinations of the nicotine content in mg. If the results are as shown below, (a) are the two labs measuring the same thing? (b) Justify the assumptions, if any, of your analytic method.

(a) Ho: μ1= μ2

Ha: μ1≠ μ2

Critical value of t = 2.306 for (5-1)+ (5-1) df at α=.05 for this 2-tailed test. Since the observed t > -2.306, we cannot reject the null hypothesis.

Lab 1 Lab 2mean 24.4 27variance 5.3 8.5

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10(b). The assumption of the t test for small samples is that the variances are equal. We can test this by the F test.

Ho: Ha: The critical value of F = 6.39 with (5-1), (5-

1) df at α= .05. Since the observed F < 6.39, we cannot re

ject the null hypothesis. Thus the assumption is justified.

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11. What is power? Power is the complement of β, which mean

s it is the probability of rejecting the null hypothesis when the null

hypothesis is in fact false.

12. A metallurgist made 5 determinations of the melting point of manganese and found an average of 1268˚ and s = 2.92˚. (a) Test that the mean of this population is no more than 5˚ more than published value of 1260˚. After thinking about his measurements, the metallurgist decided that they should have a standard deviation of 2˚ or less. (b) Are his data consistent with this?

(a) Ho: μ ≤ 1265 Ha: μ > 1265 The critical value of t = 2.132 for this 1-tailed test with (5-1) df at α= .05.

Since the observed t > 2.132, we reject the null hypothesis. (b) Ho: σ2 ≤ 22

Ha: σ2 > 22

The critical value of χ2 = 9.49 for this 1-tailed test with (5-1) df at α= .05. Since the observed χ2 < 9.49, we cannot reject the null hypothesis.

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13. If alpha is .05 and we reject the null, what is the probability that we are right?

There is no probability that we are

right. After we make the decision, we are either right or wrong and we do not know which. But the data support the decision we made.

14. A researcher classified 172 children according to intelligence and apparent family income, with results as shown below. Are these variables related?

The expected values for cellij are obtained by (ni.)(n.j)/n ; the expected values for the cells are shown in parentheses. The critical value of χ2 = 9.49 with (3-1)(3-1) df at α=.05.

Since the observed χ2 > 9.49, we reject the null hypothesis and d

eclare the variables related.

Dull Average Very smart Total Very well clothed

8 (13) 32 (34) 23 (16) 63

Well clothed 14 (15)

46 (41) 15 (19) 75

Poorly clothed 13 ( 7) 16 (19) 5 ( 8) 34 Total 35 94 43 172

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15. Explain how are hypotheses are classified.

Hypotheses are either simple or composite. A simple hypothesis is a test of a particular

value of the alternative hypothesis against the value of the null hypothesis.

A composite hypothesis is a test of more

than one value of the alternative hypothesis against the value of the null hypothesis.

Note: Many of you put null and alternative, so I gave you half credit.

16. Twelve pairs of students have been matched with respect to intelligence. One member of each pair is given a practice test before a college admissions test, and the other is not. The test scores are shown below. Is the practice test effective? If so, compute a 95% confidence interval for the difference between the means. Explain why the interval did or did not cover 0.

Ho: μdiff = 0Ha: μdiff > 0

Note: This is a 1-tailed test because the question asks if the practice test is

effective, that is, is it better than no practice test.

The critical value of t = 1.796 with (12-1) df for this 1-tailed test at α = .05. Since the observed t > 1.796, we reject the null hypothesis.

Pair 1 2 3 4 5 6 7 8 9 10 11 12Stimulant

75

73

83

83

92

97

92 85 94 91 87 93

No Stim 74

72

85

81

91

96

92 83 95 89 86 93

Diff 1 1 -2 2 1 1 0 2 -1 2 1 0

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16 (b)

The 95% confidence interval for the mean difference is

The confidence interval is -0.115 < μ < 1.45 This confidence interval covers 0 even though the null

hypothesis was rejected because the test was 1-tailed and confidence intervals are always 2-sided. This is why the t value has changed from 1.796 to 2.201.

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17. Explain how are statistical tests are classified.

Statistical tests may be either 1-tailed or 2-tailed. In a 1-tailed test, the critical region of probability α is in either the right or left tail of the test statistic distribution. In a 2-tailed test, the critical region consists of both tails of the test statistic distribution, each with probability ½ α.

Note: t, χ2, and F are types of test statistics, not types of statistical tests, but I gave you half credit.

18. Given the two sets of sample results below, (a) test that the means differ by no more than 1 unit. (b) Then test the hypothesis that both samples came from populations with the same variance.

(a) Ho: μ1- μ2 ≤ 1 Ha: μ1- μ2 > 1

The critical value of t = 1.701 for (15-1) + (15-1) df in this 1-tailed te

st at α = .05. Since the observed value of t >1.701, we reject the null hypothesis.

Sample 1 Sample 2

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18 (continued)

(b) Ho: σ12 = σ2

2

Ha: σ12 ≠ σ2

2

The critical value of F = 2.44 for (15-1),(15-1) df at α= .05. Since the observed F < 2.44, we cannot

reject the null hypothesis.

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19. (a) What assumption is required for a t test? (b) How can you check this assumption?

The t test requires homogeneity of variance (equal variances) when df is small. The test is robust to this assumption for large samples.

The two variances involved in the t test can be checked for equality by using an F test.

20. The progeny of a certain mating were classified by an attribute into three groups, as shown below. According to the genetic model, the frequencies should be in the ratio p2 : 2pq : q2. Are the data consistent with the model if p = .3?

Ho: p=.3 Ha: p ≠.3

The critical value of χ2 for (3-1) df with α=.05 is 5.99. Since the observed χ2 < 5.99, we cannot reject the null hypothesis.

Group 1 Group 2 Group 3 TotalObserved

10 53 46 109

Expected

.32(109) =10

.42(109)=46

.49(109)=53

109

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21. Why do we always put the larger variance over the smaller variance for the F test?. What about the left tail?

The two tails of the F distribution are

reciprocals of one another. So if we test by putting the larger variance over the smaller variance, we use the right tail and use F as the critical value of the test statistic.

This is equivalent to putting the smaller variance over the larger and using 1/F as the critical value of the test statistic for the left tail.

22. What is significance testing and how does it differ from hypothesis testing?

Significance testing is equivalent to hypothesis testing in all respects except that the size α of the critical region is not chosen beforehand, as it is in hypothesis testing.

Instead it is chosen by the probability of a value as large as or larger than the value of the observed test statistic.