Question

a simple random sample of size n is drawn from a population that is normally distributed. the sample mean, x, is found to be 106, and the sample standard deviation, s, is found to be 10
(a) construct a 98% confidence interval about μ if the sample size, n, is 23
(b) construct a 98% confidence interval about μ if the sample size, n, is 19
(c) construct a 96% confidence interval about μ if the sample size, n, is 23
(d) could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?
click the icon to view the table of areas under the t - distribution
(a) construct a 98% confidence interval about μ if the sample size, n, is 23
lower bound \\( \square \\), upper bound \\( \square \\)
(use ascending order. round to one decimal place as needed.)

Explanation:

Step1: Determine the critical value

The confidence level is \(98\%\), so \(\alpha = 1 - 0.98=0.02\). The degrees of freedom \(df=n - 1=23 - 1 = 22\). Using the t - distribution table, \(t_{\alpha/2}=t_{0.01}\approx 2.508\)

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\). Given \(s = 10\), \(n = 23\), then \(E=2.508\times\frac{10}{\sqrt{23}}\approx2.508\times2.062\approx5.2\)

Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\). Given \(\bar{x}=106\), then the lower bound is \(106 - 5.2 = 100.8\) and the upper bound is \(106+5.2 = 111.2\)

Answer:

Lower bound \(100.8\), Upper bound \(111.2\)