Question

An educational psychologist at a large university wants to estimate the mean IQ of the students in tertiary institutes in Macau. A random sample of 15 students (in universities in Macau) gives the following data on IQs. From further investigation, the data are normally distributed.

113	120	103	118	104	98	110	126
88	110	126	112	120	101	115

a. At the 5% significance level, do the data provide sufficient evidence to conclude that the IQ of tertiary students in Macau is higher than the common Macau adults?

b. Obtain the 95% confidence interval for the mean IQ of tertiary students in Macau.

(Given that IQs of Macau adults are normally distributed with µ = 100 and σ = 16.)

Answer 1

a)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 100
Alternative Hypothesis, Ha: μ > 100

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (110.93 - 100)/(16/sqrt(15))
z = 2.65

P-value Approach
P-value = 0.004
As P-value < 0.05, reject the null hypothesis.

the data provide sufficient evidence to conclude that the IQ of tertiary students in Macau is higher than the common Macau adults

b)
sample mean, xbar = 110.93
sample standard deviation, σ = 16
sample size, n = 15

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (110.93 - 1.96 * 16/sqrt(15) , 110.93 + 1.96 * 16/sqrt(15))
CI = (102.83 , 119.03)