Question

A random sample of 21 college? men's basketball games during the last season had an average attendance of 5,114 with a sample standard deviation of 1,795. Complete parts a and b below.

a. Construct a 90?% confidence interval to estimate the average attendance of a college? men's basketball game during the last season.The 90?% 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 size is larger than 30.

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

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

Answer 1

Solution :

Given that,

= 5114

s = 1795

n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,20 =1.724718=1.725 (rounder)

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

because random sample less than 30

correct option:D. The only assumption needed is that the population follows the? Student's t-distribution

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

= 1.725 * (1795/ 21) = 675.68

The 90% confidence interval estimate of the population mean is,

- E < < + E

5114 - 675.68 < < 5114+ 675.68

4438.32 < < 5789.68

(The 90?% confidence interval to estimate the average attendance of a college? men's basketball game during the last season is from a lower limit of 4438.3 to an upper limit of 5789.7