Question

A psychologist is interested in the mean IQ score of a given group of children. It is known that the IQ scores of the group have a standard deviation of 10. The psychologist randomly selects 80 children from this group and finds that their mean IQ score is 111. Based on this sample, find a 95% confidence interval for the true mean IQ score for all children of this group. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answer to one decimal place. ( If necessary, consult a list of formulas.)

What is the lower limit of the 95% confidence interval?

What is the upper limit of the 95% confidence interval?

Answer 1

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96 ( Critical value from Z table )
111 ± Z (0.05/2 ) * 10/√(80)
Lower Limit = 111 - Z(0.05/2) 10/√(80)
Lower Limit = 108.8087 ≈ 108.8
Upper Limit = 111 + Z(0.05/2) 10/√(80)
Upper Limit = 113.1913 ≈ 113.2
95% Confidence interval is ( 108.8 , 113.2 )