To test the belief that sons are taller than their fathers, a student randomly s

Question

To test the belief that sons are taller than their fathers, a student randomly selects 6 fathers who have adult male children. She records the height of both the father and son in inches and obtains the accompanying data. Are sons taller than their fathers? Use the alpha = 0 1 level of significance Note that a normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Click the icon to view the height data. Choose the correct null and alternative hypotheses below. Let d_i = Y_i - X_i. A. H_0: mu_d > 0 H_1: mu_d

Explanation / Answer

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d < 0
Alternative hypothesis: d > 0 , sons are taller than fathers. (Our claim)
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic.

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(3.942/6) + (2.022/6)] = 1.81
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (3.942/6 + 2.022/6)2 / { [ (3.942 / 6)2 / (5) ] + [ (2.022 / 6)2 / (5) ] }
DF = 10.675 / (1.339 + 0.092) = 7.46

t = [ (x1 - x2) - d ] / SE = [ (70.52 - 69.75) - 0 ] / 1.81 = 0.425

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a one-tailed test, the P-value is

We use the t Distribution Calculator to find P(t < 0.425) = 0.341422

Interpret results. Since the P-value (0.341422) is more than the significance level (0.10), we can accept the null hypothesis.

Conclusion. Do not reject Ho . There is not sufficient evidence to conclude that sons are taller than their fathers.

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