# Solved: Body mass index is calculated by dividing a person’s weight by the square of her height; it is a measure of the extent t

#### ByRaju Chaudhari

Oct 15, 2020

Body mass index is calculated by dividing a person's weight by the square of her height; it is a measure of the extent to which the individual overweight. For the population of middle aged men who later develop diabetes mellitis, the distribution of baseline body mass indices is approximately normal with an unknown mean $\mu$ and a standard deviation $\sigma$. A sample of 58 men selected from this group has $\overline{x}=25.0 kg/m^2$ and standard deviation $s=2.7 kg/m^2$.

(a) Construct a 95% confidence interval for the population mean $\mu$.
(b) At the 0.05 level of significance, test whether the mean baseline body mass index for the population of middle aged men who do develop diabetes is equal to 24.0kg/m2, the mean for the population of men who do not. What is the p-value of the test?
(c) What do you conclude?
(d) Base on the 95% confidence interval, would you have expected to reject or not to reject the null hypothesis? Why?

#### Solution

(a) 95% Confidence interval for the population mean $\mu$ :

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 $n = 58$, sample mean $\overline{X}= 25 kg/m^2$, sample standard deviation $s = 2.7 kg/m^2$.

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 $t_{\alpha/2,n-1} = t_{0.025,58-1}= 2.002$.

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}}\\ & = 2.002 \frac{2.7}{\sqrt{58}} \\ & = 0.71. \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\\ 25 - 0.71 & \leq \mu \leq 25 + 0.71\\ 24.29 &\leq \mu \leq 25.71. \end{aligned}
Thus, $95$% confidence interval estimate for population mean is $(24.29,25.71)$.

(b) The hypothesis testing problem is $H_0:\mu=24.0 kg/m^2$ against $H_a:\mu \neq 24.0 kg/m^2$.

Given that the sample size $n = 58$, sample mean $\overline{x}= 25/mm3$ and sample standard deviation $s = 2.7 / mm3$.

Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 24$ against $H_1 : \mu \neq 24$ ($\text{two-tailed}$)

Step 2 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.

Step 3 Significance Level

The significance level is $\alpha = 0.05$.

Step 4 Critical Value(s)

As the alternative hypothesis is $\text{two-tailed}$, the critical value of $t$ $\text{are}$ $-2.002 and 2.002$.

The rejection region (i.e. critical region) is $\text{t < -2.002 or t > 2.002}$.

Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{25-24}{2.7/ \sqrt{58 }}\\ &= 2.821 \end{aligned}

Step 6 Decision

The test statistic is $t =2.821$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

OR

Step 6 Decision

$p$-value Approach

This is a $\text{two-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=2.821$). Thus the p-value = $0.0066$.

The p-value is $0.0066$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

(c) As the p-value is less than $0.05$, there is sufficient evidence to conclude that the mean baseline body mass index for the population of middle aged men who do develop diabetes is not equal to $24.0kg/m^2$

(d) $95$% confidence interval estimate for population mean is $(24.29,25.71)$. As the claimed value $\mu = 24.0$ is not in the confidence interval, we reject the null hypothesis at $0.05$ level of significance. We get the same conclusion as obtained in part (b).