Question

IQ in appellation is normally distributed .

We know 95 % confidence interval for this population is (70.6 , 129.4 )

What does this interval represent ?

find the mean and std ofIq in this population.

hint z0.05\2 = 1.96

What is the 99% confidence interval?

hint z00.1/2 = 2.58

Answer 1

this interval represent we are 95% confident that the true mean IQ lies in between 70.6 , and 129.4.

Given  95 % confidence interval for this population is (70.6 , 129.4 )

xbar-ME=70.6

xbar+ME=129.4

adding 2 eq

2xbar=70.6+129.4

xbar=70.6+129.4/2

xbar=200/2=100

sample mean=100

we have xbar+ME=129.4

ME=129.4-100=29.4

Margin of error=zcrit*sd=29.4/1.96=15

mean and std of Iq in this population. is 100 and 15

99% confidence interval is

xbar-Z*sd,xbar+Z*sd

100-2.58*15,100+2.58*15

61.3,138.7

99% confidence interval is 61.3 to 138.7

we are 99% confident that the true mean IQ lies in between  61.3 and 138.7