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.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
