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 $(27.929,35.404)$.