a statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. the 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. suppose the distribution of the population of number of pushups that can be done is approximately normal. what is the standard error of the mean?

One day, a teacher of statistics randomly selected 10 pupils' names from a hat and instructed them to complete as many pushups as they could before class began. The 10 students who were chosen at random performed an average of 15 pushups, with a standard deviation of 9. Assume that the population's distribution of the maximum number of pushups is roughly normal.

We have to find what is the standard error of the mean.

We know that,

For this problem, we have the standard deviation for the sample, thus, the t-distribution is used.

The sample mean is of 15 is x=15 .

The sample standard deviation is of 9 is s=9.

The sample size is of 10 is n=10.

First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.

Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622

The margin of error is of:

M=ts/n

Then:

M=2.2622(9/√10)

M=6

The confidence interval is:

x±M

x-M=15-6=9

x+M=15+6=21

Therefore, Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).

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