Question

A random sample of 15 college​ men's basketball games during the last season had an average attendance of 5,154 with a sample standard deviation of 1,775. 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 normal 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​ Student's t-distribution.

Answer 1

Given, a random sample of 15 college men's basketball games during the last season had an average of 5154 with a sample standard deviation of 1775.

Let X be the random variable denoting the number of attendance. Let X₁, X₂,..., X₁₅ be the random sample from this population and let be the sample mean.

Here, we assume, X follows normal distribution. Here, both mean and variance of X are unknown. Thus, we estimate mean and variance by =5154 and s=1775.

a)

The 99% confidence interval of average attendance is,

[-(1775*t_0.005,13) /√15, +(1775*t_0.005,13) /√15]

Or, [3773, 6535] (from t-distribution table, t_{0. 005,13}=3.0123)

Thus, the 99% C. I for average attendance is [3773, 6535].

b)

Here, the only assumption is, the population follows the normal distribution.