Question

a simple random sample of size n is drawn. the sample mean, \\( \overline { x } \\), is found to be 17.6, and the sample standard deviation, s, is found to be 4.9
click the icon to view the table of areas under the t - distribution
(a) construct a 95% confidence interval about \\( \mu \\) if the sample size, n, is 35
lower bound 15.92, upper bound: 19.28
(use ascending order. round to two decimal places as needed)
(b) construct a 95% confidence interval about \\( \mu \\) if the sample size, n, is 61
lower bound: 16.35, upper bound: 18.85
(use ascending order. round to two decimal places as needed)
how does increasing the sample size affect the margin of error, e?
a. the margin of error does not change
b. the margin of error decreases
c. the margin of error increases
(c) construct a 99% confidence interval about \\( \mu \\) if the sample size, n, is 35
lower bound \\( \square \\), upper bound \\( \square \\)
(use ascending order. round to two decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom and t - value

For a sample size \(n = 35\), the degrees of freedom \(df=n - 1=35 - 1 = 34\). For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99 = 0.01\), and \(\frac{\alpha}{2}=0.005\). Looking up the t - value in the t - distribution table (or using a calculator), \(t_{\frac{\alpha}{2},df}\approx 2.728\)

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

The formula for the margin of error for a confidence interval for the population mean when the population standard deviation \(\sigma\) is unknown is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\). Given \(s = 4.9\), \(n = 35\), and \(t_{\frac{\alpha}{2}}\approx2.728\)

$$ E=2.728\times\frac{4.9}{\sqrt{35}}\approx2.728\times\frac{4.9}{5.916}\approx2.728\times0.828\approx2.26 $$

Step3: Calculate the lower and upper bounds

The sample mean \(\bar{x}=17.6\)
The lower bound \(=\bar{x}-E=17.6 - 2.26=15.34\)
The upper bound \(=\bar{x}+E=17.6+ 2.26=19.86\)

Answer:

Lower bound: \(15.34\), Upper bound: \(19.86\)