Question

The business college computing center wants to determine the proportion of business students who have personal computers (PCʹs) at home. If the proportion exceeds 25%, then the lab will scale back a proposed enlargement of its facilities. Suppose 200 business students were randomly sampled and 65 have PCʹs at home. Find the rejection region for this test using α = 0.01.

Answer 1

Answer with explanation:

By considering the given information, we have

$H_0: \mu\leq0.25\\\\H_a:\mu>0.25$, since the alternative hypothesis is right tail, so the test is a right tail test.

Given : Sample size : n= 200

Number of students have PC's in their home  = 65

Then, $\hat{p}=\dfrac{65}{200}=0.325$

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

$z=\dfrac{0.325-0.25}{\sqrt{\dfrac{0.25(0.75)}{200}}}\\\\\Rightarrow\ z= 2.45$

Using the standard normal distribution table for z, we have

P-value for right tail test  : $P(z>2.45)=1-P(z<2.45)=1-0.9928572=0.0071428$

Since the p-value is less that the significance level ($\alpha=0.01$), so we reject the null hypothesis.

Hence, we have evidence to reject the null hypothesis i.e. we may support the alternative hypothesis that the proportion exceeds 25%.

Answer 2

The probability shows that the rejection region for this test using α = 0.01 is when the null hypothesis is greater than 2.33

How to compute the probability?

From the information given, the proportion exceeds 25%, then the lab will scale back a proposed enlargement of its facilities and 200 business students were randomly sampled and 65 have PCʹs at home.

In this case, the rejection region for this test using α = 0.01 is when the null hypothesis is greater than 2.33 since the critical value is 2.33.

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