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 = [tex] \frac{z_{\alpha/2} standard deviation}{\sqrt{n}} [/tex]
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 = [tex] \frac{z_{\alpha/2} standard deviation}{\sqrt{n}} [/tex]
= [tex] \frac{2.575*10}{\sqrt{22}} [/tex]
= 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)
