Question

17) Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. Heights of men aged 25 to 34 have a standard deviation of 2.9. Use a 0.05 significance level to test the claim that the heights of women aged 25 to 34 have a different standard deviation. The heights (in inches) of 16 randomly selected women aged 25 to 34 are listed below. Round the sample standard deviation to five decimal places 62.13 65.09 64.18 66.72 63.09 61.15 67.50 64.65 63.80 64.21 60.17 68.28 66.49 62.10 65.73 64.72

Answer 1

Given Data: 62.13, 65.09, 64.18, 66.72, 63.09, 61.15, 67.5, 64.65, 63.8, 64.21, 60.17, 68.28, 66.49, 62.1, 65.73, 64.72 We calculate sample mean and std deviation from given data. Sample Mean, x = sigma (x)/n = 1030.01/16 = 64.375625 Sample Variance, s^2 = 77.874194/15 = 5.191613 Sample std dev, s = squareroot s^2 = squareroot 5.191613 almostequalto 2.278511 We want to test two tailed hypothesis: H_0: sigma = 2.9 (Null Hypothesis) H_a: sigma notequalto 2.9 (Alternative Hypothesis, also called H_1) Given: s = 2.278 51 Rightarrow s^2 = 5.191608, n = 16 Significance level, alpha = 0.05 Degrees of freedom, df = n - 1 = 15 Since this is two tailed test, we need area of alpha = 0.05 on either side of critical value. This means area of 0.025 on right of positive critical value. Critical values arc given by chi_1 - alpha/2^2 = chi_0.975^2 and chi_1 - alpha/2^2 = chi_0.975^2 15 degrees of freedom. We can use chi-square table or following excel commands: = 5.191613 Sample std dev, s = squareroot s^2 = squareroot 5.191613 almostequalto 2.278511 We want to test two tailed hypothesis: H_0: sigma = 2.9 (Null Hypothesis) H_a: sigma notequalto 2.9 (Alternative Hypothesis, also called H_1) Given: s = 2