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Anastaziya [24]
3 years ago
13

In Pennsylvania the average IQ score is 101.5 The variable is normally distributed and the population standard deviation is know

n 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
Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

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.

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