Question

In a survey conducted by a website, employers were asked if they had ever sent an employee home because they were dressed inappropriately. A total of 2759 employers responded to the survey, with 974 saying that they had sent an employee home for inappropriate attire. In a press release, the website makes the claim that more than one-third of employers have sent an employee home to change clothes.
Do the sample data provide convincing evidence in support of this claim? Test the relevant hypotheses using
α = 0.05.
For purposes of this exercise, assume that it is reasonable to regard the sample as representative of employers in the United States. (Round your test statistic to two decimal places and your P-value to four decimal places.)

Answer 1

Answer:

$z=\frac{0.353 -0.333}{\sqrt{\frac{0.333(1-0.333)}{2759}}}=2.23$  

$p_v =P(z>2.23)=0.0129$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.

Step-by-step explanation:

Data given and notation

n=2759 represent the random sample taken

X=974 represent the people saying that they had sent an employee home for inappropriate attire

$\hat p=\frac{974}{2759}=0.353$ estimated proportion of people saying that they had sent an employee home for inappropriate attire

$p_o=0.333$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 more than one-third of employers have sent an employee home to change clothes.:  

Null hypothesis:$p \leq 0.33$  

Alternative hypothesis:$p > 0.33$  

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:  

$z=\frac{0.353 -0.333}{\sqrt{\frac{0.333(1-0.333)}{2759}}}=2.23$  

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 significance level provided $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a right tailed the p value would be:  

$p_v =P(z>2.23)=0.0129$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.