Question

In Pennsylvania the average IQ score is 101.5 The variable is normally distributed and the population standard deviation is known to be 15. A school superintendent claims that students in her school district have an IQ that is significantly greater than the state average. A random sample of 30 students from her district are selected and the sample mean IQ score was 106.4 Find the p-value for the test and round to 3 decimal places.
a. 0.0736
b. 0.0465
c. 0.0368
d. 0.05
e. 0.963

Answer 1

Answer:

P-value = 0.0368

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101.5

Sample mean, $\bar{x}$ = 106.4

Sample size, n = 30

Alpha, α = 0.05

Population standard deviation, σ = 15

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 101.5\\H_A: \mu> 101.5$

We use One-tailed(right) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

Since,  

$z_{stat} > z_{critical}$

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.