Sample mean greater than 8.25

Question

Suppose it is known that the mean utility rate for the entire population is actually 7.59 cents, but that the variance is unknown. What is the probability of obtaining a sample mean greater than 8.25 cents if a random sample of 35 cities is selected and the sample has a standard deviation of 2.15 cents?

 

Needs to be done in excel dont know what formual(s) to use to get the correct answer thats all i was provided with.

Answer 1

Hi Justin,

 

When we're dealing with samples, we need to compute the standard error. The standard error is the standard deviation for the distribution of sample means. We can compute it with the following formula:

 

SE = S/√n, where S is the standard deviation and n is the sample size. In your case, this is 2.15/√35 = 0.363.

Next, we have to compute a one sample t value. This will be equal to (M - μ)/SE, where M is the sample mean and μ is the population mean. In your case, this is (8.25 - 7.59)/.363 = 1.816.

 

Now we look this value up in a t distribution table (or, in excel, we can use the t.dist function). Since we have 35 cities in the sample, we have 35 - 1 = 34 degrees of freedom. When we look this up in the table, we find that approximately 3.9% of cities have scores above 8.25.

 

Hope that helps