What is the square root of 4

According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets retu

jeka57 [31]

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%.

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