# Solved (Free): Percentages of ideal body weight were determined for 18 randomly selected insulin dependent diabetics and are shown below

#### ByDr. Raju Chaudhari

Mar 30, 2021

Percentages of ideal body weight were determined for 18 randomly selected insulin dependent diabetics and are shown below: A percentage of 120 means that an individual weighs 20% more than his or her ideal body weight; a percentage of 95 means that the individual weighs 5% less than the ideal.

107 119 99 114 120 104 88 114 124
116 101 121 152 100 125 114 95 117

Compare a two sided 95% confidence interval for the true mean percentage of ideal body weight for the population of insulin dependent diabetics.

#### Solution

Given that sample size $n = 18$.
The sample mean is

 \begin{aligned} \overline{X} &= \frac{1}{n}\sum_{i=1}^n X_i\\ &= \frac{1}{18}(2030)\\ &= 112.778 \end{aligned}

The sample standard deviation

 \begin{aligned} s & = \sqrt{\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X})^2}\\ &= \sqrt{\frac{1}{18 -1}(3537.1111)}\\ &=14.424 \end{aligned}

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 $n =18$, sample mean $\overline{X}=112.778$, sample standard deviation $s=14.424$.

#### 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 $t_{\alpha/2,n-1} = t_{0.025,18-1}= 2.11$.

#### 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}}\\ & = 2.11 \frac{14.424}{\sqrt{18}} \\ & = 7.174. \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\\ 112.778 - 7.174 & \leq \mu \leq 112.778 + 7.174\\ 105.604 &\leq \mu \leq 119.952. \end{aligned}
Thus, $95$% confidence interval for the true mean percentage of ideal body weight for the population of insulin dependent diabetics is $(105.604,119.952)$.