Question

 A guidance counselor at a certain school claims that the students in his school are above average intelligence. A random sample of thirty-five students IQ scores have a mean score of 113.5. Is there sufficient evidence to support the principal’s claim? The mean population IQ is 100 with a standard deviation of 18. Use a=0.05 (right-tailed). QUESTION: What is the critical value based on the z table: ____ 

Answer 1

Here,

Population Mean $\mu$ = 100

Population SD $\sigma =18$

Sample size n = 35

Sample mean $\bar x=113.5$

Null Hypothesis $H_o: \mu=100$

(the students have average IQ)

Alternate Hypothesis $H_1:\mu > 100$

(the students have above average IQ scores)

Let the significance level α = 0.05. The table value Zα = 1.645, since this is a one-tailed test. 

Test Statistic:

$Z=\dfrac{\bar x-\mu}{\frac{\sigma}{\sqrt n}}\\\Rightarrow \dfrac{113.5-100}{\frac{18}{\sqrt{35}}}\\\Rightarrow \dfrac{13.5}{3.04}=4.44$

Since 4.44 > 1.645

$\Rightarrow$ $Z>Z_{\alpha}$ at 5% level, we reject the null hypothesis. Hence we conclude that the students have above average IQ scores. So the principal’s claim is right.