Question

The null and alternative hypotheses for a population proportion, as well as the sample results, are given. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information.

Hypotheses: H0: p = 0.5 vs Ha: p<⁢0.5 ;
Sample data: p^ = 38/100 = 0.38 with n = 100.

Answer 1

`z=\frac{0.38 -0.5}{\sqrt{(0.5(1-0.5))/(100)}}=-2.4`

`p_v =P(z<-2.4)=0.0082`

And we can use the following excel code to find it:

"=NORM.DIST(-2.4,0,1,TRUE)"

Explanation:

Data given and notation

n=100 represent the random sample taken

`\hat p=0.38` estimated proportion of interesst

`p_o=0.5` 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is lower than 0.5:

Null hypothesis:
`p \geq 0.5`

Alternative hypothesis:
`p < 0.5`

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

`z=\frac{\hat p -p_o}{\sqrt{(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`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.38 -0.5}{\sqrt{(0.5(1-0.5))/(100)}}=-2.4`

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 left tailed test the p value would be:

`p_v =P(z<-2.4)=0.0082`

And we can use the following excel code to find it:

"=NORM.DIST(-2.4,0,1,TRUE)"