Questions: 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 60.064 .565 .063 .063 .0 Daughter 60.067 .065 .563 .063 .5 In this example, μd 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? H0: μd = 0. H1: μd ≠ 0.

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 60.064 .565 .063 .063 .0 Daughter 60.067 .065 .563 .063 .5

In this example, μd 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?

H0: μd = 0. H1: μd ≠ 0.

Transcript text: 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 60.064 .565 .063 .063 .0 Daughter 60.067 .065 .563 .063 .5 In this example, $\mu_{\mathrm{d}}$ 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? \[ \begin{array}{l} \mathrm{H}_{0}: \mu_{\mathrm{d}} \square \nabla \square \mathrm{in} . \\ \mathrm{H}_{1} \mu_{\mathrm{d}} \square \nabla \square \mathrm{in} . \end{array} \]

Solution

Solution Steps

Step 1: Calculate the Differences

For each pair of observations, we calculate the difference (e.g., daughter's height minus mother's height). Differences: [0.003, -0.5, 0.5, 0, 0.5]

Step 2: Compute Descriptive Statistics

Mean of differences ($\bar{d}$): 0.101 Standard deviation of differences (s): 0.418 Number of pairs (n): 5

Step 3: Determine the Test Statistic

Using the formula for the t-statistic: \[t = \frac{\bar{d} - \mu_0}{s/\sqrt{n}}\] t-statistic: 0.538

Step 4: Find the P-value

Using the calculated t-statistic to find the P-value. P-value: 0.619

Step 5: Make a Decision

Compare the P-value to the significance level ($\alpha$). Since the P-value 0.619 is greater than $\alpha = 0.01$, we do not reject the null hypothesis.