Solved (Free): You take a sample of 22 from a population of test scores, and the mean of your sample is 60

ByDr. Raju Chaudhari

Mar 14, 2021

You take a sample of 22 from a population of test scores, and the mean of your sample is 60.

(a) You know the standard deviation of the population is 10.What is the 99% confidence interval on the population mean.
(b) Now assume that you do not know the population standard deviation, but the standard deviation in your sample is 10. What is the 99% confidence interval on the mean now?

Solution

a. 99% confidence interval on the population mean when population standard deviation is 10:

Given that sample size $n = 22$, sample mean $\overline{X}= 60$ and population standard deviation $\sigma = 10$.

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 $n =22$, sample mean $\overline{X}=60$ and population standard deviation is $\sigma = 10$.

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 $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, 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}{\sqrt{n}}\\ & = 2.58 \frac{10}{\sqrt{22}} \\ & = 5.492. \end{aligned}

Step 6 Determine the confidence interval

$99$% confidence interval estimate for population mean is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 60 - 5.492 & \leq \mu \leq 60 + 5.492\\ 54.508 & \leq \mu \leq 65.492. \end{aligned}

Thus, $99$% confidence interval to estimate the mean number of pieces per package for the population is $(54.508,65.492)$.

b. 99% confidence interval on the population mean when sample standard deviation is 10:

Given that sample size $n = 22$, sample mean $\overline{X}= 60$, sample standard deviation $s = 10$.

The confidence level is $1-\alpha = 0.99$.

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 $n =22$, sample mean $\overline{X}=60$, sample standard deviation $s=10$.

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 $t_{\alpha/2,n-1} = t_{0.005,22-1}= 2.831$.

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}}\\ & = 2.831 \frac{10}{\sqrt{22}} \\ & = 6.036. \end{aligned}

Step 6 Determine the confidence interval

$99$% confidence interval estimate for population mean is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 60 - 6.036 & \leq \mu \leq 60 + 6.036\\ 53.964 &\leq \mu \leq 66.036. \end{aligned}
Thus, $99$% confidence interval estimate for population mean is $(53.964,66.036)$.