Question

Jerry's Fish Shack claims that 16% of its employees are late to work once a week. The manager surveyed 25 employees and found that actually 20% are late to work once a week. What does the test statistic calculation look like in this situation for an appropriate hypothesis test?

Answer 1

Answer:

We need to conduct a hypothesis in order to test the claim that the true proportion of employess are late to work is 0.16, so we need to apply a one sample proportion test:  

Null hypothesis:$p=0.16$  

Alternative hypothesis:$p \neq 0.16$  

$z=\frac{0.20 -0.16}{\sqrt{\frac{0.16(1-0.16)}{25}}}=0.546$  

$p_v =2*P(z>0.546)=0.585$  

Step-by-step explanation:

Data given and notation

n=25 represent the random sample taken

$\hat p=0.2$ estimated proportion  

$p_o=0.16$ 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 true proportion of employess are late to work is 0.16, so we need to apply a one sample proportion test:  

Null hypothesis:$p=0.16$  

Alternative hypothesis:$p \neq 0.16$  

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:  

$z=\frac{0.20 -0.16}{\sqrt{\frac{0.16(1-0.16)}{25}}}=0.546$  

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:  

$p_v =2*P(z>0.546)=0.585$