A brochure inviting subscriptions for a new diet program states that the participants are expected to lose over 22 pounds in five weeks. Suppose that, from the data of the five-week weight losses of 56 participants, the sample mean and the standard deviation are found to be 23.5 and 10.2 pounds, respectively. Could the statement in the brochure be substantiated on the basis of these findings? Test with $\alpha=0.05.$

(a) Formulate the null and the alternative hypothesis.

(b) State and calculate the test statistics from the data.

(c) Determine the critical (rejection) region for this test.

(d) Draw a conclusion.

(e) Calculate the P-value.

#### Solution

Given that $n = 56$, $\overline{x}= 23.5$, $\sigma = 10.2$.

(a) The hypothesis testing problem is $H_0 : \mu \leq 22$ against $H_1 : \mu > 22$ ($\textit{right-tailed}$)

(b) The test statistic is

` $$ \begin{equation*} Z=\frac{\overline{x} -\mu}{\sigma/\sqrt{n}} \end{equation*} $$ `

The test statistic under the null hypothesis is

` $$ \begin{eqnarray*} Z&=&\frac{\overline{x} -\mu_0}{\sigma/\sqrt{n}}\\ &=& \frac{23.5-22}{10.2/\sqrt{56}}\\ &=& 1.1 \end{eqnarray*} $$ `

(c) The significance level is $\alpha = 0.05$.

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{1.64}$.

The rejection region (i.e. critical region) is $\text{Z > 1.64}$.

(d) Conclusion

The test statistic is $Z =1.1$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

(e) $p$-value

This is a $\textit{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($Z=1.1$) is p-value = $0.1356$.

The p-value is $0.1356$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.