QUESTION:

Suppose that in the population of children in the United States, the average number of minutes per day spent exercising is 45 with a standard deviation of 10. I select a random sample of 16 children and find that they spend an average of 50 minutes exercising with a standard deviation of 12. Please use this information for answer the following questions.

Average mean = 45

Standard deviation = 10

Sample size = 16

Mean Sample = 50

Standard deviation Sample = 12

Calculate two standard errors of the mean, one using the population standard deviation and the other using the sample standard deviation. Describe what each of these standard errors tells you.

To do the two standard errors of the mean, we do:

Standard error of the mean (population s.d.): 10/16 10/4 = 2.50. This tells us that the average difference between the population mean and the sample means, when the samples are randomly selected and n = 16, is 2.50 minutes of exercise per day.

Standard error of the mean (sample s.d.): 12/16 12/4 = 3.00. This tells us that the average difference between the population mean and the sample means, when the samples are randomly selected and n = 16, is 3.00 minutes of exercise per day.

Using the standard error of the mean that you calculated using the population standard deviation, determine the probability of getting a difference between the sample mean and the population mean that is this large by chance. (Note: This is a z score problem)

I did this:

Here, we apply another z score formula:

We would do Z = Sample mean - Population mean / standard error using the population standard deviation

z = (50 - 45)/2.50 5/2.50 = 2.00

Now find the p associated with 2: .9772

p = 1 - .9772 = .0228.

Am I correct in this?

Also, can you explain why I should do 1 - the p value?