Answer:
P-value = 0.0368
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 101.5
Sample mean, [tex]\bar{x}[/tex] = 106.4
Sample size, n = 30
Alpha, α = 0.05
Population standard deviation, σ = 15
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 101.5\\H_A: \mu> 101.5[/tex]
We use One-tailed(right) z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis. Thus, students in Pennsylvania have an IQ that is significantly greater than the state average.
P-value can be calculated with the standard normal table.
P-value = 0.0368
P-value is lower than the significance level.
