StatsHomework 7-3

Question

A simple random sample from a population with a normal distribution of

105

body temperatures has

x=98.90°F

and

s=0.63°F.

Construct

a

98​%

confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

A confidence interval is a RANGE around a descriptive statistic, so is two numbers. A confidence interval is calculated by taking a descriptive statistic (like the mean) and then adding/subtracting the standard error times the a critical value to/from this statistic.

As such, a confidence interval is centered on the descriptive statistic you are making an estimate about. In this case, the mean of 98.9 is where we begin. To construct the interval, you'll now need the standard error of the mean, which in this case is S/√N or 0.63/√105 because N is the sample size (105). Now we need a critical value. To find a critical value, you need the "level of confidence", the degrees of freedom, and a t-table.

The level of confidence here is 98%, the degrees of freedom in this problem are N - 1 or 104. Here is a link to a website that can help you look up the correct critical value:

https://www.graphpad.com/quickcalcs/statratio1/

You choose the "t-ratio" and type in 104 for DF and .02 for the "probability." This is because a 98% confidence interval means we have an alpha = .02 (100% - 98% = 2% or p =Ù .02). This will give you the correct critical value.

Now you have all you need to construct the interval, which again will two numbers:

98.9 + ((0.63/√105) * critical value) and then...

98.9 - ((0.63/√105) * critical value)

I hope that is helpful!