Answer to Question #337926 in Statistics and Probability for Rose

Question

A random sample of 60 Grade 11 students’ ages is obtained to estimate the mean ages of all Grade 11 students. Suppose the sample mean is 17.3 and the population variance is 18,

What is the point estimate of the population parameter?

Find the 95% confidence interval for the population parameter?

Find the 99% confidence interval for the population parameter?

N=10 X= 82.5; s = 3. The parent population is normally distributed. Use the 95% confidence level to find the interval estimate for µ.

Answer 1

$n = 11 \\ \bar{x} = 17.3 \\ \sigma^2 = 18 \\ \sigma = \sqrt{18}$

The point estimate of the population parameter

$\mu = 17.3$

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\dfrac{\sigma}{\sqrt{n}})$$

$$=(17.3-1.96\cdot\dfrac{\sqrt{18}}{\sqrt{11}}, 17.3+1.96\cdot\dfrac{\sqrt{18}}{\sqrt{11}})$$

$$=(14.793,19.807)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $14.793 < \mu < 19.807,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(14.793, 19.807).$

The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} = 2.5758.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\dfrac{\sigma}{\sqrt{n}})$$

$$=(17.3-2.5758\cdot\dfrac{\sqrt{18}}{\sqrt{11}}, 17.3+2.5758\cdot\dfrac{\sqrt{18}}{\sqrt{11}})$$

$$=(14.005,20.595)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $14.005 < \mu < 20.595,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(14.005, 20.595).$

The critical value for $\alpha = 0.05$ and $df=n-1=9$ is $t_c = z_{1-\alpha/2; n-1} =2.262156$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\dfrac{s}{\sqrt{n}})$$

$$=(82.5-2.262156\cdot\dfrac{3}{\sqrt{10}}, 82.5+2.262156\cdot\dfrac{3}{\sqrt{10}})$$

$$=(80.354,84.646)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $80.354< \mu < 84.646,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(80.354,84.646).$