# Solved: Is the Diet Practical? When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and

#### ByDr. Raju Chaudhari

Oct 9, 2020

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 lb. Based on these results, does the diet appear to be effective? Does the diet appear to have practical significance?

### Solution

Given that the sample size $n = 40$, sample mean $\overline{x}= 3$ and sample standard deviation $s = 4.9$.

Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 0$ against $H_1 : \mu > 0$ ($\text{right-tailed}$)

Test Statistic

The test statistic is

 \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

Significance Level

The significance level is $\alpha = 0.01$.

Critical Value

As the alternative hypothesis is $\text{right-tailed}$, the critical value of $t$ for $39$ degrees of freedom $\text{is}$ $2.426$.

That is $P(t_{39}> t_{crit})=0.01 \Rightarrow t_{crit} =2.426$.

The rejection region (i.e. critical region) is $\text{t > 2.426}$.

Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{3-0}{4.9/ \sqrt{40 }}\\ &= 3.872 \end{aligned}

Decision

The test statistic is $t =3.872$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

$p$-value Approach

This is a $\text{right-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=3.872$) is p-value = $P(t> 3.872)=0.0002$.

The p-value is $0.0002$ which is $\text{less than}$ the significance level of $\alpha = 0.01$, we $\text{reject}$ the null hypothesis.

We conclude that at $0.01$ level of significance the diet appears to be effective. That is the diet appears to have practical significance.