# Solved: The table below shows the weights of seven randomly selected subjects before and after following a p

#### ByRaju Chaudhari

Oct 9, 2020

The table below shows the weights of seven randomly selected subjects before and after following a particular diet for two months. The dependent sample are from two normally distributed populations. Do the data suggest that the diet is effective in reducing weight? Use a 0.01 significance level.

Subject A B C D E F G
Before 191 185 165 191 193 176 161
After 184 176 163 196 179 178 149

### Solution

The sample size $n = 7$. Let $d=x-y$. So

Before After d d-dbar (d-dbar)^2
191 184 7 1.714286 2.938776
185 176 9 3.714286 13.795918
165 163 2 -3.285714 10.795918
191 196 -5 -10.285714 105.795918
193 179 14 8.714286 75.938776
176 178 -2 -7.285714 53.081633
161 149 12 6.714286 45.081633

 \begin{aligned} \overline{d}& = \frac{1}{n}\sum d_i\\ &= \frac{37}{7}\\ &=5.2857 \end{aligned}
and

 \begin{aligned} s_d^2 &= \frac{1}{n-1}\sum (d_i-\overline{d})^2\\ &=\frac{307.4285714}{7-1}\\ &=51.2383956 \end{aligned}
$s_d =7.1581$.

Step 1 Hypothesis

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

Step 2 Test Statistic

The test statistic is

 \begin{aligned} t=\frac{\overline{d} -\mu_d}{s_d/\sqrt{n}} \end{aligned}

Step 3 Level of Significance

The significance level is $\alpha = 0.01$.

Step 4 Critical Value(s)

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $t$ for $6$ degrees of freedom and $\alpha = 0.01$ level of significance $\text{is}$ $\text{3.143}$.

The rejection region (i.e. critical region) is $\text{t > 3.143}$.

Step 5 Computation

The test statistic for testing above hypothesis testing problem under the null hypothesis is

 \begin{aligned} t&=\frac{\overline{d} -\mu_d}{s_d/\sqrt{n}}\\ &= \frac{5.2857-0}{7.1581/\sqrt{7}}\\ &= 1.9537 \end{aligned}

Step 6 Decision

The test statistic is $t =1.9537$ which falls $\textit{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\textit{right-tailed}$ test, so p-value is the area to the $\textit{right}$ of the test statistic ($t=1.9537$). That is p-value = $P(t\geq 1.9537 ) = 0.0493$.

The p-value is $0.0493$ which is $\textit{greater than}$ the significance level
of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

There is no sufficient evidence to suggest that the diet is effective in reducing weight.