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

Answer:

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

$p_v =P(z$  

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

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

Step-by-step 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:  

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

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

Calculate the statistic  

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

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

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:  

$p_v =P(z$  

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

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