Question

According to the University of Nevada Center for Logistics Management, of all merchandise sold in the United States gets returned. A Houston department store sampled items sold in January and found that of the items were returned.

a. Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store.b. Construct a 95% confidence interval for the porportion of returns at the Houston store.c. Is the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole? Provide statistical support for your answer.

Answer 1

a)
`\hat p=(12)/(80)=0.15` estimated proportion of items that were returned

b) The 95% confidence interval would be given (0.0718;0.228).

c) Using a significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole.

Explanation:

Assuming:

According to the University of Nevada Center for Logistics Management, 6% of all mer-chandise sold in the United States gets returned. Houston department store sampled 80 items sold in January and found that 12 of the items were returned.

Data given and notation

n=80 represent the random sample taken

X=12 represent the items that were returned

`\hat p=(12)/(80)=0.15` estimated proportion of items that were returned

`\alpha=0.05` represent the significance level (no given, but is assumed)

Confidence =0.95 or 95%

p= population proportion of items that were returned

a. Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store

`\hat p=(12)/(80)=0.15` estimated proportion of items that were returned

b. Construct a 95% confidence interval for the porportion of returns at the Houston store

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`0.15 - 1.96 \sqrt{(0.15(1-0.15))/(80)}=0.0718`

`0.15 + 1.96 \sqrt{(0.15(1-0.15))/(80)}=0.228`

And the 95% confidence interval would be given (0.0718;0.228).

c. Is the proportion of returns at the Houston store significantly different from the returns for the nation as a whole? Provide statistical support for your answer.

We need to conduct a hypothesis in order to test the claim that the population proportion differs significantly to the USA proportion of 6% or no. We have the following system of hypothesis :

Null Hypothesis:
`p = 0.06`

Alternative Hypothesis:
`p \\eq 0.06`

We assume that the proportion follows a normal distribution.

This is a two tailed test for the proportion .

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 =80*0.15=12>10`

`n(1-p_o)=80*(1-0.15)=68>10`

Calculate the statistic

The statistic is calculated with the following formula:

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

On this case the value of
`p_o=0.06` is the value that we are testing and n = 80.

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

The p value for the test would be:

`p_v =2*P(z>3.390)=0.00070`

Using a significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole.