Listed in the accompanying table are heights (in .) of mothers and their first daughters. The data pairs are from a journal kept by Francis Galton. Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal Use a 0.01 significance level to test the claim that there is no difference in heights between mothers and their first daughters. Mother 64.064564 .565 .063567 .060 .0585 e Daughter $65.566567 .069064 .5620650665 t$ In this example, $\mu_{d}$ is the mean value of the differences $d$ for the population of all pairs of data, where each individual difference $d$ is defined as the daughter's height minus the mother's height. What are the null and alternative hypotheses for the hypothesis test? \[ \begin{array}{l} H_{0} \cdot \mu_{\mathrm{d}} \nabla \square \text { in. } \\ H_{1}, \mu_{\mathrm{d}} \nabla \square \text { in. } \end{array} \] (Type integers or decimals. Do not round)
Solution
Step 1 ：In this example, \(\mu_{d}\) is the mean value of the differences \(d\) for the population of all pairs of data, where each individual difference \(d\) is defined as the daughter's height minus the mother's height.
Step 2 ：The null hypothesis (H0) is usually a statement of no effect or no difference. The alternative hypothesis (H1) is the statement that there is an effect or difference.
Step 3 ：In this case, the null hypothesis would be that there is no difference in heights between mothers and their first daughters, which means the mean difference (\(\mu_{d}\)) is 0.
Step 4 ：The alternative hypothesis would be that there is a difference in heights between mothers and their first daughters, which means the mean difference (\(\mu_{d}\)) is not 0.
Step 5 ：\(\boxed{\text{The null hypothesis, } H0: \mu_{d} = 0 \text{ in.}}\)
Step 6 ：\(\boxed{\text{The alternative hypothesis, } H1: \mu_{d} \neq 0 \text{ in.}}\)
