Question

A random sample of 19 college​ men's basketball games during the last season had an average attendance of 5,045 with a sample standard deviation of 1,754.

Construct a 90% confidence interval to estimate the average attendance of a college​ men's basketball game during the last season.

My answer was wrong 4,842.78 and 5,247.22

The correct answer is: 4,430 and 5,660

Please explain step by step.

Answer 1

Sample mean is $\bar X$ = 5045

sample standard deviation is s = 1754.

Sample size, n = 19 and the required confidence level is 90%.

degrees of freedom, df = 19 - 1 = 18, and the significance level is $\alpha$ = 0.1.

Based on the provided information, the critical t-value for $\alpha$ = 0.1 and df = 18 of freedom is t_c = 1.73

The 90% confidence for the population mean is computed using the following expression

$CI = \left(\bar X - \frac{t_c \times s}{\sqrt{n}} , \bar X + \frac{t_c \times s}{\sqrt{n}} \right)$

Therefore, based on the information provided, the 90% confidence for the population mean is

$CI = \left(5045 - \frac{ 1.734 \times 1754}{\sqrt{ 19}} ,\ 5045 + \frac{ 1.734 \times 1754}{\sqrt{ 19}} \right)$

$= (5045 - 697.753, 5045 + 697.753)$

$=(\textbf{4347.247, 5742.753})$

which completes the calculation.

This is the correct answer.