Question

The business college computing center wants to determine the proportion of business students who have personal computers (PC's) at home. If the proportion differs from 25%, then the lab will modify a proposed enlargement of its facilities. Suppose a hypothesis test is conducted and the test statistic is 2.4. Find the P-value for a two-tailed test of hypothesis.

Answer 1

Answer:

$p_v =2*P(Z>2.4)=0.016$  

And we can use the following excel code:

"=2*(1-NORM.DIST(2.4;0;1;TRUE))"

Step-by-step explanation:

1) Data given and notation

n represent the random sample taken

X represent the business students who have personal computers (PC's) at home

$\hat p$ estimated proportion of business students who have personal computers (PC's) at home

$p_o=0.25$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.25:  

Null hypothesis:$p=0.25$  

Alternative hypothesis:$p \neq 0.25$  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic  

For this case the value of the statistic is given by z=2.4 and that's given.

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(Z>2.4)=0.016$  

And we can use the following excel code:

"=2*(1-NORM.DIST(2.4;0;1;TRUE))"