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: 100.8; upper bound: 111.2
(use ascending order. round to one decimal place as needed.)
(b) construct a 98% confidence interval about μ if the sample size, n, is 19
lower bound: ; upper bound:
(use ascending order. round to one decimal place as needed.)

Explanation:

Step1: Find the critical value \(t_{\alpha/2}\)

For a 98% confidence interval, \(\alpha = 1 - 0.98=0.02\), and \(\alpha/2 = 0.01\). The degrees of freedom \(df=n - 1=19 - 1 = 18\).
From the t - distribution table, \(t_{0.01,18}=2.552\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is unknown is \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\).
Given \(\bar{x}=106\), \(s = 10\), \(n = 19\).
\(E=2.552\times\frac{10}{\sqrt{19}}\approx2.552\times2.294\approx5.876\)

Step3: Calculate the lower and upper bounds

The lower bound \(=\bar{x}-E=106 - 5.876\approx100.1\)
The upper bound \(=\bar{x}+E=106 + 5.876\approx111.9\)

Answer:

Lower bound: \(100.1\); Upper bound: \(111.9\)