The "Freshman 15 " refers to the belief that college students gain $15 \mathrm{lb}$ (or $6.8 \mathrm{~kg}$ ) during their freshman year. Listed in the accompanying table are weights $(\mathrm{kg})$ of randomly selected male college freshmen. The weights were measured in September and later in April. 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. Complete parts (a) through (c). September $\quad 5671706460 \quad 5068 \quad 6464$ April $\begin{array}{llllllllll}58 & 77 & 68 & 66 & 64 & 47 & 69 & 68 & 60\end{array}$ a. Use a u.us signiticance level to test the ciaim tnat tor the population or tresnman male coliege students, tne weignts in September are less than the weights in the following April. In this example, $\mu_{\mathrm{d}}$ is the mean value of the differences $\mathrm{d}$ for the population of all pairs of data, where each individual difference $d$ is defined as the April weight minus the September weight. What are the null and alternative hypotheses for the hypothesis test? \[ \begin{array}{l} H_{0}: \mu_{d}=0 \mathrm{~kg} \\ H_{1}: \mu_{d}>0 \mathrm{~kg} \end{array} \] (Type integers or decimals. Do not round.) Identify the test statistic. $t=0.97$ (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 question is asking to perform a hypothesis test to determine if the weights of male college freshmen in September are less than the weights in the following April. The null hypothesis is that the mean difference in weights is zero, and the alternative hypothesis is that the mean difference is greater than zero. The test statistic has already been provided as 0.97.
Step 2 ：The P-value is the probability of obtaining a result as extreme as the observed result, under the assumption that the null hypothesis is true. In this case, we are performing a one-tailed test, so we need to find the probability of obtaining a test statistic as extreme as 0.97 under the null hypothesis.
Step 3 ：Using the given test statistic of 0.97 and degrees of freedom of 8, we calculate the P-value.
Step 4 ：The P-value is approximately 0.180. This is the probability of obtaining a test statistic as extreme as 0.97 under the null hypothesis.
Step 5 ：Since the P-value is greater than the significance level of 0.05, we do not reject the null hypothesis. This means that we do not have sufficient evidence to support the claim that the weights of male college freshmen in September are less than the weights in the following April.
Step 6 ：Final Answer: The P-value is approximately \(\boxed{0.180}\).
