Respuesta :
Question:
A simple random sample with n=54 provided a sample mean of 22.5 and a sample standard deviation of 4.4.
a. Develop a 90% confidence interval for the population mean.
b. Develop a 95% confidence interval for the population mean.
c. Develop a 99% confidence interval for the population mean.
d. What happens to the margin of error and the confidence interval as the confidence level is increased?
Answer:
a. CI = 21.4976 ≤ μ ≤ 23.5024
b. CI = 21.29903 ≤ μ ≤ 23.70097
c. CI = 20.90021 ≤ μ ≤ 24.09979
d. The margin of error and confidence interval increases
Step-by-step explanation:
Here we have
Sample size, n = 54
Mean, [tex]\bar{x}[/tex] = 22.5
s = 4.4
a. The confidence interval, CI is;
[tex]CI=\bar{x}\pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}[/tex]
At c = 90% confidence level, [tex]t_{\alpha/2}[/tex] with degrees of freedom, df = n - 1 = 54 - 1 = 53 and α = (1 - c)/2 = (1 - 0.9)/2 = 0.05, from the t table or relation we will find
[tex]t_{\alpha/2}[/tex] = 1.674116
Therefore, plugging in the values, we have
CI at 90% gives
[tex]CI=22.5\pm 1.674 \times \frac{4.4}{\sqrt{54}}[/tex]
or CI = 21.4976 ≤ μ ≤ 23.5024
b. For c = 95%, α = (1 - 0.95)/2 = 0.025, df = 54 - 1 = 53
From tables or t relations, [tex]t_{\alpha/2}[/tex] = 2.005746
∴ [tex]CI=22.5\pm 2.005746 \times \frac{4.4}{\sqrt{54}}[/tex]
or CI = 21.29903 ≤ μ ≤ 23.70097
c. For c = 99%, α = (1 - 0.99)/2 = 0.005, df = 54 - 1 = 53
From tables or t relations, [tex]t_{\alpha/2}[/tex] = 2.671822
∴ [tex]CI=22.5\pm 2.671822\times \frac{4.4}{\sqrt{54}}[/tex]
or CI = 20.90021 ≤ μ ≤ 24.09979
d. As the confidence level is increased, [tex]t_{\alpha/2}[/tex] increases, therefore, the margin of error and confidence interval increases.
