Question

Nicki's business ships orders from a location in Chicago and another location in Denver. She suspected that a larger proportion of orders shipped from the Denver location were arriving late. She took a random sample of 100 orders from each location. She found that 20 orders in the Denver sample arrived late, and 12 orders in the Chicago sample arrived late. She wants to use these results to test if the proportion of orders that arrive late is significantly larger at the Denver location. She will test H, PDenver = PChicago versus H, : PDenver > Pohicago. Assume that all conditions have been met. What is the P-value associated with these sample results? What is the P-value associated with these sample results?

Choose 1 answer
A P-value < 0.01
B 0.01 < P-value < 0.05
c.0.05 < P-value < 0.10
d.0.10 < P-value < 0.15
E P-value > 0.15

Answer 1

To test if a greater proportion of orders from Denver arrive late compared to Chicago, a two-proportion z-test is applied. The p-value calculated using the 2-PropZTest function is 0.0417, which suggests at the 5% significance level, one would reject the null hypothesis that the proportions are equal.

Step-by-step explanation:

To calculate the p-value for Nicki's suspicion that a larger proportion of orders shipped from Denver are arriving late compared to those from Chicago, we can use a two-proportion z-test since we have two independent samples. The hypothesis being tested is: H0: PDenver = PChicago against the alternative hypothesis Ha: PDenver > PChicago. Given that the samples are large enough (both > 5), we assume the sampling distribution of the proportion is approximately normally distributed, which allows us to use this kind of z-test.

To proceed with the test, we first find the z-value using the formula for two proportions, followed by finding the corresponding p-value based on the normal distribution. Since the calculated p-value determines if we reject or fail to reject the null hypothesis, it's crucial to compare it with the significance level, usually denoted as alpha (α).

We are given several p-values and related information in the provided references. Most closely, we can see that using a calculator function 2-PropZTest, the p-value = 0.0417. If we compare this with typical significance levels such as 0.05, 0.10, or 0.01, we can see that this p-value would indicate we reject the null hypothesis at the 5% significance level (since 0.0417 < 0.05), but not at the 1% significance level (since 0.0417 > 0.01). Given the p-values in the original question options, the correct answer fits within the B: 0.01 < P-value < 0.05 range.