# Solved (Free): The distribution of systolic and diastolic blood pressure for female diabetes between the ages of 30 and 34 have unknown means

#### ByDr. Raju Chaudhari

Apr 12, 2021

The distribution of systolic and diastolic blood pressure for female diabetes between the ages of 30 and 34 have unknown means. However the standard deviation are $\sigma_s=11.8$ mm Hg and $\sigma_d =9.1$ mm Hg, respectively

a) A random sample of ten women is selected from this population. The mean systolic blood pressure for the sample is =130 mm Hg. Calculate a two sided 95% confidence interval for $\mu_s$, the true mean systolic blood pressure.
b) Interpret this confidence interval
c) The mean diastolic pressure for the sample of size 10 is =84 mm Hg. Find two sided 90% confidence interval for $\mu_d$, the true mean diastolic blood pressure of the population.
d) Calculate a two-sided 99% confidence interval for $\mu_d$.
e) How does the 99% confidence interval compare to the 90% interval?

#### Solution

a. Two sided 95% confidence interval for $\mu_s$, the true mean systolic blood pressure

##### 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 $n =10$, sample mean $\overline{X}=130$ and population standard deviation is $\sigma = 11.8$.

##### Step 3 Specify the formula

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

 \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned}
where $E = Z_{\alpha/2} \frac{\sigma_s}{\sqrt{n}}$, and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

##### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

##### Step 5 Compute the margin of error

The margin of error for mean is

 \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma_s}{\sqrt{n}}\\ & = 1.96 \frac{11.8}{\sqrt{10}} \\ & = 7.314. \end{aligned}

##### Step 6 Determine the confidence interval

$95$ % confidence interval estimate for population mean $\mu_s$ is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 130 - 7.314 & \leq \mu \leq 130 + 7.314\\ 122.686 & \leq \mu \leq 137.314. \end{aligned}

Thus, $95$% confidence interval for population mean $\mu_s$ is $(122.686 mm Hg,137.314 mm Hg)$.

b. We are 95% confident that the interval $(122.686 mm Hg,137.314 mm Hg)$ contains the true mean for systolic blood pressure $\mu_s$.

c. Two sided 90% confidence interval for $\mu_d$, the true mean diastolic blood pressure

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

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

##### Step 2 Given information

Given that sample size $n =10$, sample mean $\overline{X}=84$ and population standard deviation is $\sigma_d = 9.1$.

##### Step 3 Specify the formula

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

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

where $E = Z_{\alpha/2} \frac{\sigma_d}{\sqrt{n}}$, and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

##### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.05} = 1.65$.

##### Step 5 Compute the margin of error

The margin of error for mean is
 \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma_d}{\sqrt{n}}\\ & = 1.65 \frac{9.1}{\sqrt{10}} \\ & = 4.734. \end{aligned}

##### Step 6 Determine the confidence interval

$90$ % confidence interval estimate for population mean $\mu_d$ is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 84 - 4.734 & \leq \mu \leq 84 + 4.734\\ 79.266 & \leq \mu \leq 88.734. \end{aligned}

Thus, $90$% confidence interval for population mean $\mu_d$ is $(79.266 mm Hg,88.734 mm Hg)$.

d. Two sided 99% confidence interval for $\mu_d$, the true mean diastolic blood pressure

##### 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

Given that sample size $n =10$, sample mean $\overline{X}=84$ and population standard deviation is $\sigma_d = 9.1$.

##### Step 3 Specify the formula

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

 \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned}
where $E = Z_{\alpha/2} \frac{\sigma_d}{\sqrt{n}}$, and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

##### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.005} = 2.58$.

##### Step 5 Compute the margin of error

The margin of error for mean is
 \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma_d}{\sqrt{n}}\\ & = 2.58 \frac{9.1}{\sqrt{10}} \\ & = 7.413. \end{aligned}

##### Step 6 Determine the confidence interval

$99$ % confidence interval estimate for population mean $\mu_d$ is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 84 - 7.413 & \leq \mu \leq 84 + 7.413\\ 76.587 & \leq \mu \leq 91.413. \end{aligned}

Thus, $99$% confidence interval for population mean $\mu_d$ is $(76.587 mm Hg,91.413 mm Hg)$.

e. The 99% confidence interval for $\mu_d$ is wider than the 90 % confidence interval for $\mu_d$. As the confidence coefficient increases the confidence interval becomes wider.