A random sample of 200 adults was given an IQ test It was fo

A random sample of 200 adults was given an IQ test. It was found that 92 of them scored higher than 100. Based on this, compute a 95% confidence interval for the proportion of all adults whose IQ score is greater than 100.

Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.

Q.1 - What is the lower limit of the 95% confidence interval?

Q.2 - What is the upper limit of the 95% confidence interval?

Solution

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=92
Sample Size(n)=100
Sample proportion = x/n =0.92
Confidence Interval = [ 0.92 ±Z a/2 ( Sqrt ( 0.92*0.08) /100)]
= [ 0.92 - 1.96* Sqrt(0) , 0.92 + 1.96* Sqrt(0) ]
= [ 0.87,0.97]