Question

A random sample of 18 college? men's basketball games during the last season had an average attendance of 5,038 with a sample standard deviation of 1,766.

Complete parts a and b below.

a. Construct a 99?% confidence interval to estimate the average attendance of a college? men's basketball game during the last season.

The 99?% confidence interval to estimate the average attendance of a college? men's basketball game during the last season is from a lower limit of __ to an upper limit of __.

?(Round to the nearest whole? numbers.)

b. What assumptions need to be made about this? population?

A. The only assumption needed is that the population distribution is skewed to one side.

B. The only assumption needed is that the population follows the? Student's t-distribution.

C. The only assumption needed is that the population size is larger than 30.

D. The only assumption needed is that the population follows the normal distribution.

Answer 1

Solution :

Given that,

=5038

s = 1766

n = 18

freedom = df = n - 1 = 18- 1 = 17

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,17 = 2.89823=2.898 (rounded)

Margin of error = E = t_/2,df * (s /n)   

= 2.898* (1766 / 18) = 1206.29

The 99% confidence interval estimate of the population mean is,

- E < < + E

5038 - 1206.29 < < 5038 + 1206.29

3831.71 < < 6244.29

(3831.71, 6244.29 )

The 99?% confidence interval to estimate the average attendance of a college? men's basketball game during the last season is from a lower limit of 3831.7 to an upper limit of 6244.3

(b)The only assumption needed is that the population follows the? Student's t-distribution

because sample size less than 30