Question

According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets returned. A Houston department store sampled 80 items sold in January and found that 12 of the items returned. Is there evidence to show the proportion of returns at the Houston store was more than the national expectation at the .10 level of significance? Compute the p-value and interpret its meaning. Explain your conclusions.

Answer 1

Answer:

$p_v =P(z>3.390)=0.000349$  

If we compare the p value and the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of returns at the Houston store is significantly higher than 0.06 or 6%.

Step-by-step explanation:

1) Data given and notation n  

n=80 represent the random sample taken

X=12 represent number of items returned

$\hat p=\frac{12}{80}=0.15$ estimated proportion of items returned

$p_o=0.06$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that proportion of returns at the Houston store was more than the national expectation (0.06):  

Null hypothesis:$p\leq 0.06$  

Alternative hypothesis:$p >0.06$  

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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.15 -0.06}{\sqrt{\frac{0.06(1-0.06)}{80}}}=3.390$  

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 significance level provided $\alpha=0.1$. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

$p_v =P(z>3.390)=0.000349$  

If we compare the p value and the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of returns at the Houston store is significantly higher than 0.06 or 6%.