Question

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

Answer 1

Answer:

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$, 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 $t_{\alpha/2}=2.33$

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

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$    

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

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.