Question

According to a research center, 6% of all merchandise sold in a particular country gets returned. A department store in a certain city sampled 80 items sold in January and found that 12 of the items were returned.

A. Construct a point estimate of the proportion of items returned for the population of sales transactions at the store in the given city.

B. Construct a 95% confidence interval for the proportion of returns at the store in the given city. (Round your answers to four decimal places.)

C. Is the proportion of returns at the store in the given city significantly different from the returns for the country as a whole? Provide statistical support for your answer.

Develop appropriate hypotheses such that rejection of H₀ will support the conclusion that the proportion of returns at the store in the given city is significantly different from the returns for the country as a whole.

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to four decimal places.)

At α = 0.01, what is your conclusion?

Answer 1

a)

Point estimate = sample proportion, = 0.15

b)

sample size, n = 80
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.15 * (1 - 0.15)/80) = 0.03992

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.03992
ME = 0.07824

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.15 - 1.96 * 0.03992 , 0.15 + 1.96 * 0.03992)
CI = (0.0718 , 0.2282)

c)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.06
Alternative Hypothesis, Ha: p ≠ 0.06

Rejection Region
This is two tailed test, for α = 0.01
Critical value of z are -2.58 and 2.58.
Hence reject H0 if z < -2.58 or z > 2.58

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.15 - 0.06)/sqrt(0.06*(1-0.06)/80)
z = 3.39

P-value Approach
P-value = 0.0007
As P-value < 0.01, reject the null hypothesis.