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Question 7, 9.3.13-T
Part 2 of 4
HW Score: $44.97 \%, 4.5$ of 10 points
Points: 0 of 1
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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.
\[
62.066 .563 .069 .063 .567 .064 .065 .062 .064 .5 \text { 막 }
\]
\[
65.068 .563 .069 .066 .067 .065 .565 .568 .066 .5
\]
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.
\[
\mathrm{t}=
\]
(Round to two decimal places as needed.)

Step 1 ：The null and alternative hypotheses for the hypothesis test are \(H_{0}: \mu_{d}=0 \text{ in. }\) and \(H_{1}: \mu_{d} \neq 0 \text{ in. }\) respectively.

Step 2 ：The test statistic for a t-test is calculated as follows: \(t = \frac{\text{sample mean} - \text{population mean}}{\text{sample standard deviation} / \sqrt{\text{sample size}}}\). In this case, the population mean (under the null hypothesis) is 0.

Step 3 ：The heights of mothers are [6.2066e+01, 5.6300e-01, 6.9000e-02, 6.3000e-02, 5.6700e-01, 6.4000e-02, 6.5000e-02, 6.2000e-02, 6.4000e-02, 5.0000e-01] and the heights of daughters are [6.5068e+01, 5.6300e-01, 6.9000e-02, 6.6000e-02, 6.7000e-02, 6.5000e-02, 5.6500e-01, 5.6800e-01, 6.6000e-02, 5.0000e-01].

Step 4 ：The differences between the heights of mothers and daughters are [3.002e+00, 0.000e+00, 0.000e+00, 3.000e-03, -5.000e-01, 1.000e-03, 5.000e-01, 5.060e-01, 2.000e-03, 0.000e+00].

Step 5 ：The sample mean of the differences is 0.3513999999999996, the sample standard deviation is 0.9737233465186879, and the sample size is 10.

Step 6 ：Substituting these values into the formula for the t-test statistic, we get \(t = \frac{0.3513999999999996 - 0}{0.9737233465186879 / \sqrt{10}} = 1.1412115913170626\).

Step 7 ：The test statistic for the hypothesis test is approximately \(\boxed{1.14}\).