Question

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Is the Diet Practical? When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb (based on data from “Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction,” by Dansinger et al., Journal of the American Medical Association, Vol. 293, No. 1). Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. Based on these results, does the diet appear to have statistical significance? Does the diet appear to have practical significance?

Short answer

The hypotheses are as follows.

$$\begin{gathered} H_0:\mu =0 \\ {H}_1:\mu >0 \end{gathered}$$

The test statistic is $t=3.872$, and the critical value is $2.426$.

In this case, reject $H_0$.Therefore, there is sufficient evidence that the mean weight loss is greater than 0.

The diet appears to have statistical significance, but due to a small weight loss mean of 3.0 lb, the diet does not have any practical significance.

Step-by-step solution

Given information

The summary for the weight loss for the weight watchers’ diet is as follows.

Sample size $n=40$.

Sample mean $\overline{x}=3.0\text{lb}$.

Sample standard deviation $s=4.9\text{lb}$.

The level of significance $\alpha =0.01$.

State the hypothesis

Null hypothesis: The mean weight loss is equal to 0.

$H_0:\mu =0$

Alternative hypothesis: The mean weight loss is greater than 0.

$H_0:\mu >0$

The test is right-tailed.

Compute the test statistic

As the sample is selected by random sampling along with a normally distributed population, the requirements of the t-test are satisfied. For an unknown population standard deviation, the t-test is appropriate.

The test statistic is given as follows.

$$\begin{gathered} t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}} \\ =\frac{3-0}{\frac{4.9}{\sqrt{40}}} \\ =3.872 \end{gathered}$$

.

Compute the critical value

The critical value is obtained with a significance level of 0.01, and the degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ =40-1 \\ =39 \end{gathered}$$

Refer to the t-table for the degrees of freedom 39 and the level of significance 0.01 for the one-tailed test to obtain the critical value 2.426$\left(t_{0.01}\right)$.

State the decision rule

The decision rule states the following:

If the test statistic is greater than the critical value, the null hypothesis will be rejected.

If the test statistic is not greater than the critical value, the null hypothesis will fail to be rejected.

Here, the test statistic 3.872 is greater than the critical value of 2.426.

Thus, the null hypothesis will be rejected.

Thus, it can be concluded that there is sufficient evidence to support the claim that the mean weight loss is greater than 0.

Discuss the significance of the result

The result is statistically significant as the null hypothesis is rejected at a 0.01 level of significance. It implies that the diet has a statistically significant effect on weight loss among subjects.

On the other hand, the diet does not have any practical significance as the mean weight loss after the diet is 3.0 lb, which is not very high.