Question

a. compute the sample mean and sample standard deviation for private and public colleges. round your answers to two decimal places.
\\( \overline { x } _ { 1 } = 42.49 \\)
\\( \overline { x } _ { 2 } = 22.17 \\)
\\( s _ { 1 } = 7.11 \\)
\\( s _ { 2 } = 4.51 \\)
b. what is the point estimate of the difference between the two population means? round your answer to one decimal place.
20.3
interpret this value in terms of the annual cost of attending private and public colleges.
the mean annual cost to attend private colleges is \\( \\$ 20.3 \\) more
than the mean annual cost to attend public colleges.
c. develop a \\( 95 \\% \\) confidence interval of the difference between the mean annual cost of attending private and public colleges.
\\( 95 \\% \\) confidence interval, private colleges have a population mean annual cost \\( \\$ \\) to \\$ more expensive than public colleges.

Explanation:

Step1: Determine the formula for the confidence interval

For the difference between two means \(\mu_1-\mu_2\) (where \(\mu_1\) is the population mean of private colleges and \(\mu_2\) is the population mean of public colleges), when the population variances are unknown, the formula for the confidence interval is \((\bar{x}_1 - \bar{x}_2)\pm t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\)

Assume \(n_1 = 8\) (number of private - college data points) and \(n_2=8\) (number of public - college data points). The degrees of freedom \(df=\min(n_1 - 1,n_2 - 1)=7\). For a \(95\%\) confidence interval, \(\alpha=0.05\) and \(\alpha/2 = 0.025\). From the \(t\) - distribution table, \(t_{0.025,7}=2.365\)

We know that \(\bar{x}_1 = 42.49\), \(s_1 = 7.11\), \(\bar{x}_2=22.17\), \(s_2 = 4.51\)

Step2: Calculate the margin of error

First, calculate \(\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=\sqrt{\frac{7.11^{2}}{8}+\frac{4.51^{2}}{8}}\)

$$
LATEXBLOCK0
$$

\(\sqrt{8.861525}\approx2.98\)

The margin of error \(E=t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=2.365\times2.98\approx7.05\)

Step3: Calculate the confidence interval

The lower limit is \((\bar{x}_1-\bar{x}_2)-E=(42.49 - 22.17)-7.05=13.27\)

The upper limit is \((\bar{x}_1-\bar{x}_2)+E=(42.49 - 22.17)+7.05=27.37\)

Answer:

The \(95\%\) confidence interval is \(\$13.27\) to \(\$27.37\)