1. A psychologist would like to estimate the average IQ of Canadians. (IQ scores are known...

1. A psychologist would like to estimate the average IQ of Canadians. (IQ scores are known to follow a normal distribution with standard deviation 15.) She takes a sample of 49 Canadians and measures a sample mean IQ of 107.34 and a sample standard deviation of 17.36.

(a) Use the information above to calculate a 92% confidence interval to estimate the average IQ of Canadians.

(b) Interpret the confidence interval obtained in part (a).

at the 0.08 significance level. Is this possible to do with the confidence interval calculated in part (a)? If so, what is the correct conclusion? If not, why not?

(d) Suppose the researcher also wishes to test the hypotheses
H0 :μ=105 vs. μ>105

at the 0.08 significance level. Is this possible to do with the confidence interval calculated in part (a)? If so, what is the correct conclusion? If not, why not?

Solutions

Expert Solution

a) population std dev =15 which is known hence we perform one sample Z-test.

One-Sample Z

Descriptive Statistics

N	Mean	SE Mean	92% CI for μ
49	107.34	2.14	(103.589, 111.091)

μ: mean of Sample
Known standard deviation = 15

b) We are 92 % confident that the population mean of IQ scores for canadians will lie in between (103.589,111.091)

H0 :μ=103 vs. μ̸=103

0.08 significance level.

Yes it is possible to do with the confidence interval calculated in part (a) if the given mean is in the CI then we fail to reject null hypothesis otherwise reject null hypothesis.

Since μ=103 does not present in 92% CI we reject null hypothesis and conclude that there is a significant evidence that population mean is different from 103.

H0 :μ=105 vs. μ>105

Yes it is possible to do with the confidence interval calculated in part (a) if the given mean is in the CI then we fail to reject null hypothesis other =wise reject null hypothesis.

Since μ=105 is present in 92% CI we fail to reject null hypothesis and conclude that there is no significant evidence that population mean is greater than 105.

orchestra answered 9 months ago

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