Question

The dean of students of a large community college claims that the average distance that commuting students travel to the campus is 32 miles. The commuting students feel otherwise. A sample of 64 students was randomly selected and yielded a mean of 35 miles and a standard deviation of 5 miles. Find and interpret the 90% confidence interval for the average distance that commuting students travel to the campus.

Answer 1

Answer:(33.922, 36.078)

Interpretation : We are 90% confident that the true average distance that commuting students travel to the campus lies between 33.922 and 36.078.

Step-by-step explanation:

Formula to find the confidence interval for population mean (when population standard deviation is unknown):-

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

, where n = sample size.

$\overline{x}$ = sample mean.

$s$ = sample standard deviation.

$t_{\alpha/2}$ = two tailed t-value respected to the given confidence level (or significance level) and degree of freedom.

Let x represents the distance that commuting students travel to the campus .

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ n= 60

Degree of freedom(df) = 59    [df= n-1]

$\overline{x}=35$

$s=5$

here, population standard deviation is not given , so we use t-test.

Confidence level : 90%

Significance level : $\alpha=1-0.90=0.10$

Critical value for Significance level of 0.10 and degree of freedom 63 :-

$t_{{df, \alpha/2}}=t_{(63,0.05)}=1.6694$

Then , the  90% confidence interval for the average distance that commuting students travel to the campus :-

$35\pm (1.6694)\dfrac{5}{\sqrt{60}}\\\\ 35\pm 1.078\\\\=(35- 1.078,\ 35+ 1.078)\\\\=(33.922,\ 36.078)$

Thus , the 90% confidence interval for the average distance that commuting students travel to the campus : (33.922, 36.078)

Interpretation : We are 90% confident that the true average distance that commuting students travel to the campus lies between 33.922 and 36.078.