Question

Research is being conducted to determine if a large supermarket is discriminating against African Americans. For the supermarket, thirty percent of the employees that are eligible for a merit raise are African American. Since the implement of this new merit raise program 17 out of 60 employees that received the merit raise are African American. Is there evidence to support a claim of discrimination based on race (the proportion of African Americans who receive the merit raise is less that it should be)?How should we write the alternative hypothesis?A. Ha: p < 0.3 B. Ha: p > 0.3 C. Ha: p-hat > 0.3 D. Ha: p-hat < 0.3

Answer 1

Answer:

A. Ha: p < 0.3

Using the significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we have enough evidence at this significance level to FAIL to reject the null hypothesis. And on this case we can say that the proportion of employees that received the merit raise are African American  is not significantly less than 0.3 or 30%.

Step-by-step explanation:

1) Data given and notation

n=60 represent the random sample taken  

X=17 represent the employees that received the merit raise are African American

$\hat p=\frac{17}{60}=0.283$ estimated proportion of employees that received the merit raise are African American

$p_o=0.3$ is the value that we want to test  

$\alpha$ represent the significance level (no given)  

z would represent the statistic (variable of interest)  

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

p= population proportion of employees that received the merit raise are African American

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of employees that received the merit raise are African American is less than 0.3 or 30%. :  

Null Hypothesis: $p \geq 0.3$

Alternative Hypothesis: $p$

A. Ha: p < 0.3

We assume that the proportion follows a normal distribution.  

This is a one tail lower test for the proportion of employees that received the merit raise are African American .

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =60*0.3=18>10$

$n(1-p_o)=60*(1-0.3)=42>10$

3) Calculate the statistic  

The statistic is calculated with the following formula:

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

On this case the value of $p_o=0.3$ is the value that we are testing and n = 60.

$z=\frac{0.283 -0.3}{\sqrt{\frac{0.3(1-0.3)}{60}}}=-0.287$

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.  

Based on the alternative hypothesis the p value would be given by:

$p_v =P(Z$  

Using the significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we have enough evidence at this significance level to FAIL to reject the null hypothesis. And on this case we can say that the proportion of employees that received the merit raise are African American  is not significantly less than 0.3 or 30%.