Question

Problem Page 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 16 . The psychologist randomly selects 60 children from this group and finds that their mean IQ score is 106 . 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 answers 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

Solution :

Given that,

Point estimate = sample mean = = 106

sample standard deviation = s = 16

sample size = n = 60

Degrees of freedom = df = n - 1 = 59

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,59 = 2.001

Margin of error = E = t_/2,df * (s /n)

= 2.001 * (16 / 60)

= 4.133

The 95% confidence interval estimate of the population mean is,

- E < < + E

106 - 4.133 < < 106 + 4.133

101.9 < < 110.1

The lower limit of the 95 % confidence interval is 101.9

The upper limit of the 95 % confidence interval is 110.1