Question 1161497: a. The average IQ is 100, with a population standard deviation of 15. The mayor of Smartsville claims that citizens in his town have higher than average IQs. To test that claim, 40 Smartsville citizens were randomly selected, and their average IQ was determined to be 104.1. Does this support the mayor’s claim? Use α = 0.1.
b. construct a 90 % confidence interval for the mean. Does the confidence interval support the claim made in part a?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a z-test with the usual assumptions
z=(x bar-mean)/sigma/sqrt (n)
Critical value is |z|>1.645
z=(104.1-100)*sqrt(40)/15
=1.73
reject Ho and the average IQ is not 100 in Smartsville, it is higher, p-value is 0.084
90% CI half-interval is z*sigma/sqrt(40) with z=1.645
half-interval is 3.90
add and subtract from/to mean
90% Confidence interval is (100.2, 108), and because 100 is not in the interval, is consistent with rejection of Ho in a.
|
|
|
