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

Sample size $n=40$.

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

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

The level of significance $伪=0.01$.

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

$H_0:渭=0$

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

$H_0:渭>0$

The test is right-tailed.

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.

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

.

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$\left(t_{0.01}\right)$.

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.

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.