PSTAT 5E Practice Homework 7 Solutions ) ( )

8.1 In this exercise, the parameter of interest is µ, the population mean. The objective of the experiment is to show that the mean exceed 2.3. a. We want to prove the alternative hypothesis that µ is, in fact, greater than 2.3. Hence, the alternative hypothesis is Ha : µ > 2.3 and the null hypothesis is H0 : µ = 2.3 b. the probability of a Type I error is defined as = P[reject H0 when H0 is true] = P[decide µ > 2.3 when µ = 2.3] c. The best estimator for µ is the sample average x , and the test statistics is x−µ z=

σ

n which represent the distance (measured in units if standard deviations) from x to the hypothesized mean µ.

In order to perform the test at significance level =.05, given the hypotheses above, we complete the following steps: 2.4 − 2.3 (i) The value of the test statistic is z* = = 2.04 0.29 / 35 (ii) P-value = P(Z > z *) = P(Z > 2.04) = 0.0207 , which we find with the help of the Normal Table (iii) The p-value of 0.0207 is smaller than the significance level ( =.05). Therefore, we reject the null hypothesis (iv) The conclusion is that there is evidence in the sample that the population mean µ is larger than 2.3. 8.2. If the experimenter wishes to prove that µ < 2.9, this is the alternative hypothesis. Hence, the hypothesis to be tested is H0 : µ = 2.9 Ha : µ < 2.9 and the test is one-tailed. Let’s perform the test at significance level =.05. 2.4 − 2.9 (i) The value of the test statistic (the same z-statistic from 8.1), is z* = = −10.2 0.29 / 35 (ii) P-value = P(Z < −10.2) ≈ 0 (iii) The p-value of 0 is smaller than the significance level level ( =.05). Therefore, we reject the null hypothesis

(iv) The conclusion is that there is evidence in the sample that the population mean µ is smaller than 2.9. 8.3. If the experimenter wishes to prove that µ 2.9, this is the alternative hypothesis. Hence, the hypothesis to be tested is H0 : µ = 2.9 Ha : µ 2.9 and the test is two-tailed. Testing at =.05, we have the same test statistic and value of the test statistic as above (8.2). The p-value is: 2 P(Z > z * ) = 2 P(Z > 10.2) ≈ 0 The decision and conclusion remain as above. 8.9. (i) The hypothesis of interest is H0 : µ = 48 Ha : µ 48 (ii) The test statistic is x−µ 47.1 − .48 t= with value t* = = −2.076 s 4 .7 n 25 (iii) The p-value is: 2 P(t > t *) = 2 P(t > 2.07 ) ≈ 2(0.025) = 0.05 Since the t-table at the back of the book is not detailed enough, we are very often unable to determine the exact probability of a t-random variable. We can, however, estimate it (bound it from above or below), as we did above. Ask your TA or Prof. Wang if you have doubts about how to do this. (iv) The p-value of 0.05 is less than the significance level of 0.1. Therefore, we reject the null hypothesis. (v) The conclusion is that there is evidence in the sample to suggest that the population mean µ is different from 48. 8.13 (i) The test is one-tailed with H0 : µ = 1200 Ha : µ < 1200 (ii) The observed value of the test statistic is x − µ 1186 − 1200 z* = ≈ = −1.54 σ 2480 n 30 (iii) The p-value is P(Z < −1.54) = 0.0618 from the Normal Table (iv) The p-value of 0.0618 is smaller than the significance level of 0.1. Therefore, we can reject the null hypothesis (v) The conclusion is that there is evidence that the population mean is smaller than 1200. 8.14. (i) The hypothesis to be tested is H0 : µ = 30.31 Ha : µ < 30.31 (ii) Calculate xi =578.7 and xi2 =18,462.09. Then

xi

578.7 = 28.935 n 20 ( xi ) 2 (578.7) 2 2 xi − 18,462.09 − n 20 s2 = = = 90.3898 n −1 19 The observed value of the test statistic is x − µ 28.935 − 30.31 t* = = = −.647 s 90.3898 n 20 (iii) The p-value is P(t < −0.647 ) = P(t > 0.647 ) > 0.1 , evaluating as before. (iv) Since the p-value of 0.1 is greater than 0.05, we cannot reject the null hypothesis. (v) The conclusion is that there is no evidence in the sample that the sample mean has decreased. x=

=

8.26. Note that this is a test for the difference between two means and the samples are independent. (i) The hypothesis of interest is one-tailed: H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 > 0

(ii) For this exercise, the test statistic is ( x − x 2 ) − D0 11.6 − 9.7 − 0 = 2.087 z= 1 and its value is z* = 2 2 27.9 38.4 σ1 σ 2 + + 80 80 n1 n2

(iii) The p-value is P(Z > 2.087 ) = 0.018 (iv) Since the p-value of 0.018 is smaller than 0.1, we can reject the null hypothesis. (v) The conclusion is that there is evidence to indicate that µ 1 - µ 2 > 0, or µ 1 > µ 2. 8.28. a. a. If your research objective were to show that µ 1 is different from µ 2 (whether larger or smaller), the test would be two-tailed. b. The hypothesis of interest would be (i) H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 0 (ii) The test statistic has a value z* = 2.087 (iii) The p-value is 2 P(Z > 2.087 ) = 2(0.018) = 0.036 (iv) Since the p-value is smaller than the significance level of 0.05, we reject the null hypothesis (v) There is reason to believe that µ 1 is different from µ 2. 8.30. (i) The hypothesis of interest is one-tailed: H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 > 0 (ii) The preliminary calculations are shown below: Sample 1

Sample 2

x1i = 49 2

x 1i = 617

x 2i = 38 2

x 2i = 366

x1 = 12.25

x 2 = 9.5

n1 = 4

n2 = 4

(49) 2 (38) 2 + 366 − 4 4 = 16.75 + 5.00 = 3.625 s2 = 4+4−2 6 The test statistic, under the assumption that σ 12 = σ 22 = σ 2 , is calculated using the pooled value of s 2 in the t-statistic shown below: ( x − x2 ) − 0 12.25 − 9.5 t= 1 = = 2.043 1 1 1 2 1 s ( + ) 3.625( + ) n1 n2 4 4 617 −

(iii) the p-value is P(t > 2.043) < 0.05 (iv) Since the p-value is less than the significance level, we reject the null hypothesis (v) There is evidence that µ1 > µ 2 8.76. See section 8.2 of the text 8.77. See section 8.4 of the text.

8.79. A student’s t test can be employed to test a hypothesis about a single population mean when the sample has been randomly selected from a normal population. It will work quite satisfactorily for populations that possess mound-shaped frequency distributions resembling the normal distribution. 8.80. As in the case of the single population mean, random samples must be independently drawn from two populations that process normal distributions with a common variance, σ 2 . Consequently, it is logical

that information in the two sample variance s12 and s 22 , should be pooled in order to give the best estimate of the common variance σ 2 . In this way, all of the sample information is being utilized to its best advantage. 8.82. a. If the airline is to determine whether or not the flight is unprofitable, they are interested in finding out whether or not µ < 0 (since a flight is profitable if µ is at least 60). Hence, the hypothesis to be tested is H0 : µ = 60 Ha : µ < 60 b. Since only small values of x (and hence negative values of z) would tend to disprove H0 in favor of Ha, this is a one-tailed test.

c. For this exercise, n = 120, x = 58, and s = 11. Hence, the test statistic is x − µ x − µ 58 − 60 z= ≈ = = −1.992 s 11 σ 120 n n The p-value is P(Z 6) < 0.01 Since the p-value is less than the significance level 0.05, we reject the null hypothesis. We can conclude that there is a difference in mean scores for the two methods.

d =

=

8.108. The hypothesis of interest is H0 : µ = 16 Ha : µ < 16 and the test statistic is x − µ 15.7 − 16 t= = = − 1 .8 s .5 n 9 The p-value is P (t < -1.8) = P (t > 1.8) > 0.05 and H0 is not rejected. There is insufficient evidence to indicate that the mean is less than 16.

8.110. The hypothesis of interest is H0 : µ = 1100 Ha : µ < 1100 The test statistic is x − µ 1060 − 1100 z≈ = = −1.897 s 340 n 260 The p-value is P (Z < -1.897) = 0.0294 This is less than the significance level of 0.05 and H0 is rejected. There is sufficient evidence to indicate that the mean is less than 1100; there has been a drop in average daily production.