Question

You take a sample of 22 from a population of test scores, and the mean of your sample is 60. you know the standard deviation of the population is 10. what is the 99% confidence interval on the population mean.

Answer 1

Given: sample size n=22

Sample mean m=60

Population standard deviation σ =10

As the population standard deviation is given we will use z confidence interval for population mean.

The 99% confidence interval for Population mean is given by

(Sample mean - Margin of error, Sample mean + Margin of error)

Where Sample mean m=60

Margin of error = $\frac{z_{\alpha/2} standard deviation}{\sqrt{n}}$

Where alpha = 1- c = 1- 0.99 = 0.01

z (α/2) = z (0.005)

It means area below -z and and above z is 0.005

In the z score table look for the probability value 0.005 or close to 0.005. Corresponding z score value is

In the z score table there is no accurate probability value 0.005 . We have two close values 0.0049 and 0.0051 . Value 0.005 lies in middle of these two values.

The z score value for 0.0049 is -2.58 and for 0.0051 is -2.57

S the average of -2.58 and -2.57 will be our required z score value.

z score = (-2.57) + (-2.58) / 2 = -2.575

So we get two z score values one for left size and one for right side. We will use positive z score value 2.575

The margin of error = $\frac{z_{\alpha/2} standard deviation}{\sqrt{n}}$

= $\frac{2.575*10}{\sqrt{22}}$

= 5.4899

The 99% confidence interval for population mean will be

(60 - 5.4899, 60 + 5.4899)

(54.5101, 65.4899)

The 99% confidence interval is (54.5101, 65.4899)