Question

The football coach randomly selected 10 players and timed how long each player took to perform a certain drill. The result has a sample mean of 9.48 minutes and sample standard deviation of 2.14 minutes. Round answers to two decimals. The 95% confidence interval for the mean time for all players is : __________

Answer 1

Answer:

The 95% confidence interval for the mean time for all players, in minutes, is: (7.95, 11.01).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 10 - 1 = 9

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.2622.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.2622\frac{2.14}{\sqrt{10}} = 1.53$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 9.48 - 1.53 = 7.95 minutes.

The upper end of the interval is the sample mean added to M. So it is 9.48 + 1.53 = 11.01 minutes.

The 95% confidence interval for the mean time for all players, in minutes, is: (7.95, 11.01).