# Solved:A random sample of 32 persons attending a certain diet clinic was found to have lost (over a three-week period) an average of 30 pounds

#### ByRaju Chaudhari

Sep 25, 2020

A random sample of 32 persons attending a certain diet clinic was found to have lost (over a three-week period) an average of 30 pounds, with a sample standard deviation of 11. For these data, a 99% confidence interval for the true mean weigh loss by patients attending clinic would have limits (? , ?).

#### Solution

Sample mean

The sample mean of $X$ is $\overline{x}=30\text{ pound}$.

Sample standard deviation

The sample standard deviation is $s=11 \text{ pound}$

Sample size $n = 32$, sample mean $\overline{X}= 30$, sample standard deviation $s = 11$.

The confidence level is $1-\alpha = 0.99$.

#### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

#### Step 2 Given information

Sample size $n =32$, sample mean $\overline{X}=30$, sample standard deviation $s=11$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

 \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned}

where $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$, andand $t_{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

#### Step 4 Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus $t_{\alpha/2,n-1} = t_{0.005,32-1}= 2.744$.

#### Step 5 Compute the margin of error

The margin of error for mean is
 \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.744 \frac{11}{\sqrt{32}} \\ & = 5.3358. \end{aligned}

#### Step 6 Determine the confidence interval

$99$% confidence interval estimate for population mean is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 30 - 5.336 & \leq \mu \leq 30 + 5.336\\ 24.6642 &\leq \mu \leq 35.3358. \end{aligned}

Thus, $99$% confidence interval estimate for population mean is $(24.6642,35.3358)$.

#### Interpretation

We can be $99$% confident that the true mean weigh loss by patients attending clinic would have limits $24.6642$ and $35.3358$.