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 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters. \[ \begin{array}{l} 62.066 .563 .069 .063 .567 .064 .065 .062 .064 .5 \\ 65.068 .563 .069 .066 .067 .065 .565 .568 .066 .5 \end{array} \] 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? \[ \begin{array}{l} H_{0}: \mu_{d}=0 \text { in. } \\ H_{1}: \mu_{d} \neq 0 \text { in. } \end{array} \] (Type integers or decimals. Do not round) Identify the test statistic. $\mathrm{t}=2.98$ (Round to two decimal places as needed) Identify the P-value. P-value $=\square$ (Round to three decimal places as needed.)
Solution
Step 1 ：The problem provides us with the heights of mothers and their first daughters. We are asked to test the claim that there is no difference in heights between mothers and their first daughters using a 0.05 significance level. The null hypothesis \(H_{0}: \mu_{d}=0\) in. and the alternative hypothesis \(H_{1}: \mu_{d} \neq 0\) in. are given.
Step 2 ：The test statistic for this hypothesis test is identified as \(t=2.98\).
Step 3 ：The P-value is the probability of obtaining a result as extreme as the observed result, given that the null hypothesis is true. It can be calculated using the cumulative distribution function (CDF) of the t-distribution. Since the alternative hypothesis is two-sided (the mean difference could be either positive or negative), the P-value is twice the probability that a random variable drawn from the t-distribution is greater than the absolute value of the test statistic.
Step 4 ：Given that the test statistic is 2.98 and the degrees of freedom is 18, the P-value is calculated to be 0.008026815755941108.
Step 5 ：Rounding to three decimal places as needed, the final answer is: The P-value is \(\boxed{0.008}\).
