Answer to Question #346758 in Statistics and Probability for Bethsheba Kiap

Question

A random sample of size 144 is drawn from a population whose distribution, mean, and standard deviation are all unknown. The summary statistics are 𝑥̅= 58.2 and s = 2.6.

a. Construct an 80% confidence interval for the population mean μ.

b. Construct a 90% confidence interval for the population mean μ.

Answer 1

a. The critical value for $\alpha = 0.20, df=n-1=143$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=1.2875$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(58.2-1.2875\times\dfrac{2.6}{\sqrt{144}}, 58.2+1.2875\times\dfrac{2.6}{\sqrt{144}})$$

$$=(57.921, 58.479)$$

Therefore, based on the data provided, the 80% confidence interval for the population mean is $57.291 < \mu < 58.479,$ which indicates that we are 80% confident that the true population mean $\mu$ is contained by the interval $(57.921, 58.479).$

b. The critical value for $\alpha = 0.10, df=n-1=143$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=1.655579$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$ $$=(58.2-1.655579\times\dfrac{2.6}{\sqrt{144}},$$ $$58.2+1.655579\times\dfrac{2.6}{\sqrt{144}})$$

$$=(57.841, 58.559)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $57.841 < \mu < 58.559,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(57.841, 58.559).$