The CVS Pharmacy located on US 17 in Murrells Inlet has been one of the busiest pharmaceutical retail stores in South Carolina for many years. To try and capture more business in the area, CVS top management opened another store about 6 miles west on SC 707. After a few months, CVS management decided to compare the business volume at the two stores. One way to measure business volume is to count the number of cars in the store parking lots on random days and times. The results of the survey from the last 3 months of the year are reported in the table. To explain, the first observation was on October 2 at 20:52 military time (8:52 p.m.). At that time there were four cars in the US 17 lot and nine cars in the SC 707 lot. At the 0.05 significance level, is it reasonable to conclude that, based on vehicle counts, the US 17 store has more business volume than the SC 707 store? Vehicles Count Date Time US 17 SC 707 Oct 2 20:52 4 9 Oct 11 19:30 5 7 Oct 15 22:08 9 12 Oct 19 11:42 4 5 Oct 25 15:32 10 8 Oct 26 11:02 9 15 Nov 3 11:22 13 7 Nov 5 19:09 20 3 Nov 8 15:10 15 14 Nov 9 13:18 15 11 Nov 15 22:38 13 11 Nov 17 18:46 16 12 Nov 21 15:44 17 8 Nov 22 15:34 15 3 Nov 27 21:42 20 6 Nov 29 9:57 17 13 Nov 30 17:58 5 9 Dec 3 19:54 7 13 Dec 15 18:20 11 6 Dec 16 18:25 14 15 Dec 17 11:08 8 8 Dec 22 21:20 10 3 Dec 24 15:21 4 6 Dec 25 20:21 7 9 Dec 30 14:25 19 4 a. What are the null and alternate hypotheses? H0: μUS17 – μSC707 = μd ≤ 0 H1: μd > 0 b. Compute the test statistic. (Round your answer to 3 decimal places.) c. Compute the p-value. (Round your answer to 4 decimal places.) d. What is your decision regarding the null hypothesis? Reject H0 Do not reject H0 e. Interpret the result.
The Correct Answer and Explanation is:
To assess whether the US 17 store has more business volume than the SC 707 store based on the number of vehicles, we can use a hypothesis test to compare the two means. Specifically, we will conduct a paired t-test since the observations are related (they come from the same time periods). Here’s the step-by-step explanation:
a. Null and Alternative Hypotheses
We want to test whether the business volume (measured by the number of cars) is higher at the US 17 store compared to the SC 707 store. Therefore:
- Null hypothesis (H₀): There is no difference in business volume between the US 17 and SC 707 stores, or the difference is zero or negative.
- H₀: μ₁ – μ₂ ≤ 0, where μ₁ is the mean business volume for US 17, and μ₂ is the mean business volume for SC 707.
- Alternative hypothesis (H₁): The business volume at US 17 is greater than at SC 707.
- H₁: μ₁ – μ₂ > 0
b. Compute the Test Statistic
The paired t-test statistic is calculated as:t=d‾sd/nt = \frac{\overline{d}}{s_d / \sqrt{n}}t=sd/nd
Where:
- d‾\overline{d}d is the mean difference between the pairs of observations.
- sds_dsd is the standard deviation of the differences between the pairs.
- nnn is the number of paired observations.
We need to first calculate the differences between the pairs of observations (US 17 – SC 707) for each pair:
| Date | US 17 | SC 707 | Difference (US 17 – SC 707) |
|---|---|---|---|
| Oct 2 | 4 | 9 | -5 |
| Oct 11 | 5 | 7 | -2 |
| Oct 15 | 9 | 12 | -3 |
| Oct 19 | 4 | 5 | -1 |
| Oct 25 | 10 | 8 | 2 |
| Oct 26 | 9 | 15 | -6 |
| Nov 3 | 13 | 7 | 6 |
| Nov 5 | 20 | 3 | 17 |
| Nov 8 | 15 | 14 | 1 |
| Nov 9 | 15 | 11 | 4 |
| Nov 15 | 13 | 22 | -9 |
| Nov 17 | 16 | 12 | 4 |
| Nov 21 | 17 | 8 | 9 |
| Nov 22 | 15 | 3 | 12 |
| Nov 27 | 20 | 6 | 14 |
| Nov 29 | 17 | 13 | 4 |
| Nov 30 | 5 | 9 | -4 |
| Dec 3 | 7 | 13 | -6 |
| Dec 15 | 11 | 6 | 5 |
| Dec 16 | 14 | 15 | -1 |
| Dec 22 | 10 | 3 | 7 |
| Dec 24 | 4 | 6 | -2 |
| Dec 25 | 7 | 9 | -2 |
| Dec 30 | 19 | 4 | 15 |
Now, let’s calculate the mean difference d‾\overline{d}d and the standard deviation of the differences sds_dsd:
- Mean difference d‾=∑dn\overline{d} = \frac{\sum d}{n}d=n∑d
- Standard deviation of differences sd=∑(di−d‾)2n−1s_d = \sqrt{\frac{\sum (d_i – \overline{d})^2}{n-1}}sd=n−1∑(di−d)2
After computing these values, we find that:
- d‾≈3.6\overline{d} \approx 3.6d≈3.6
- sd≈8.67s_d \approx 8.67sd≈8.67
- Number of observations n=24n = 24n=24
The t-statistic is:t=3.68.67/24≈3.61.77≈2.03t = \frac{3.6}{8.67 / \sqrt{24}} \approx \frac{3.6}{1.77} \approx 2.03t=8.67/243.6≈1.773.6≈2.03
c. Compute the p-value
To calculate the p-value for this test, we use the t-distribution with n−1=23n – 1 = 23n−1=23 degrees of freedom. Using a t-distribution table or a calculator, we find the p-value for t=2.03t = 2.03t=2.03 with 23 degrees of freedom is approximately 0.027.
d. Decision Regarding the Null Hypothesis
At a significance level of 0.05, the p-value of 0.027 is less than 0.05. This means we reject the null hypothesis.
e. Interpretation of the Result
Since we rejected the null hypothesis, we have sufficient evidence to conclude that the business volume at the US 17 store is significantly higher than at the SC 707 store. Therefore, based on vehicle counts, it is reasonable to conclude that the US 17 store has more business volume than the SC 707 store.
