Suppose that the sample standard deviation was s = 5.1. compute a 98% confidence interval for μ, the mean time spent volunteering for the population of parents of school-aged children. (round your answers to three decimal place

The 95% confidence interval would be given by (5.139;5.861)  

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Confidence interval

Assuming that and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.98 or 98%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that

Now we have everything in order to replace into formula (1):

So on this case the 95% confidence interval would be given by (5.139;5.861)    We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

im not sure

step-by-step explanation: