Brion H. answered • 08/29/23
Experienced Lean Six Sigma Tutor and Coach for Passing Exams
First, calculate the mean (average) and standard deviation of body temperature of adults in your town, following a random sampling plan.
Next, determine how many standard deviations from the mean will achieve 80% confidence interval. Typically, 95% confidence interval is used, which is about 2 standard deviations (Z-value = 1.96). So 80% will be much less than 2 standard deviations from the mean.
If 80% is inside the interval, then 20% is outside the interval. Since the normal distribution is symmetric, that equates to 10% on each side of the average. We next would look up 10% area under the curve in the Normal distribution table, to see what Z-value it provides to us. So we look up 0.10 in the following table.
https://sites.berry.edu/vbissonnette/wp-content/uploads/sites/21/2015/07/z.pdf
In this table, "Cum p" is the area under the curve to the left of the Z-value, and "Tail p" is the area to the right. Look for the Z-value with a "Tail p" = 0.10
The closest value is between 1.28 and 1.29. From experience, I know that 1.282 is the exact value, but for this example you can use 1.28 (it's closer to 0.10 than 1.29)
The formula for the 80% confidence interval = mean +/- (Z-value)*(std dev/√n)
We use +/- to look at both sides of the mean, which gets us to the 20% outside the interval, leaving the middle 80% interval.
Let's use an example:
Mean = 98.5
Std Dev = 0.3
n = 100
80% confidence interval ==> Z-value of 1.28
98.5 +/- 1.28*(0.3/√100)
98.5 +/- 1.28*(0.3/10) = 98.5 +/- 1.28*(0.03) = 98.5 +/- 0.0384
= 98.5 + 0.0384 = 98.5384
= 98.5 - 0.0384 = 98.4616
80% confidence interval would be (94.46, 98.54) to two decimal places
