# Solved (Free): The following are body mass index (BMD) scores measured in 12 patients who are free of diabetes and participating in a study of risk factors for obesity

#### ByDr. Raju Chaudhari

Mar 13, 2021

The following are body mass index (BMD) scores measured in 12 patients who are free of diabetes and participating in a study of risk factors for obesity. Body mass index is measured as the ratio of weight in kilograms to height in meters squared.

Generate a 95% confidence interval estimate of the true BMI.

25 27 31 33 26 28 38 41 24 32 35 40

#### Solution

Given that sample size $n = 12$, sample mean $\overline{X}= 31.667$, sample standard deviation $s = 5.883$.

The confidence level is $1-\alpha = 0.95$.

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

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

##### Step 2 Given information

Given that sample size $n =12$, sample mean $\overline{X}=31.6667$, sample standard deviation $s=5.883$.

##### 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 $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$, and $t_{\alpha/2, n-1}$ 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 $t_{\alpha/2,n-1} = t_{0.025,12-1}= 2.201$.

##### 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}}\\ & = 2.201 \frac{5.883}{\sqrt{12}} \\ & = 3.738. \end{aligned}

##### Step 6 Determine the confidence interval

$95$% confidence interval estimate for population mean is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 31.667 - 3.738 & \leq \mu \leq 31.667 + 3.738\\ 27.929 &\leq \mu \leq 35.404. \end{aligned}

Thus, $95$% confidence interval estimate for population mean is $(27.929,35.404)$.