Question

17.38 A and B

Exercises 409 Excercises 17.38 through 17.41 ask you to answer questions from data. As sieme that the simple conditions" hold in cach case. The exercise statements give you the State step of the fourstep process. In your work, follow the Plan, Solve, and Conclude steps, illustrated in Example 16.3 (page 383) for a confidence interval and in Example 17.7 for a test of significance. 17.38 Pulling wood apart. How heavy a load (in pounds) needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Here are data from students do ing a laboratory exercise: WOOD 33,190 31,860 32,590 26,520 33,280 32,320 33,020 32,030 30,460 32,700 23,040 30,930 32,720 33,650 32,340 24,050 30,170 31,300 28,730 31,920 We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that character- istics of materials vary Normally. Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds. (a) Is there statistically significant evidence at the a = 0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative? (b) Is there statistically significant evidence at the a = 0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative? 17.39 Bone loss by nursing mothers. As discussed in Exercise 16.26 (page 390), breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Re- searchers measured the percent change in mineral con tent of the spines of 47 mothers during three months of breast-feeding." Here are the data: S BONELS -4.7 -2.5 -4.9 -2.7 -0.8 -5.3-8.3 -2.1 -6.8 -4.3 2.2 - 7.8 -3.1 -1.0 -6.5 - 1.8 -5.2 -5.7 -7.0 -2.2 -6.5 -1.0 -3.0-3.6 -5.2 - 2.0 -2.1 -5.6 -4.4 -3.3 - 4.0 - 4.9 -4.7 -3.8 -5.9 -2.5 -0.3 -6.2 -6.8 1.7 0.3 -2.3 0.4 -5.3 0.2 -2.2 -5.1

Answer 1

SOLUTION A] (1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ=32500

Ha:μ≠​32500

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the critical value for a two-tailed test is zc​=1.64.

The rejection region for this two-tailed test is R={z:∣z∣>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

X - μo 30841 - 32500 3000/V20 = 2.473

(4) Decision about the null hypothesis: Since it is observed that ∣z∣=2.473>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0134, and since p=0.0134<0.1, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population meanμ is different than 32500, at the 0.1 significance level.

B] (1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho μ=31500

Ha:μ≠​31500

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the critical value for a two-tailed test is zc​=1.64.

The rejection region for this two-tailed test is R={z:∣z∣>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

X - μo σ//η 30841 - 31500 3000/20 = 0.982

(4) Decision about the null hypothesis: Since it is observed that z∣=0.982≤zc​=1.64, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.3259, and since p=0.3259≥0.1, it is concluded that the null hypothesis is not rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 31500, at the 0.1 significance level.