The Vancouver Chamber of Commerce claims that the mean cost of hotels in the downtown area of the city is less than in Edmonton. A random sample of 50 hotels in Vancouver (population 1) had a mean nightly rate of $109 with a standard deviation of $15. An independent sample of 60 hotels in Edmonton (population 2) had a mean nightly rate of $114 with a standard deviation of $12. In testing the hypothesis with a 5% level of significance, what is the value of the standardized test statistic and the correct conclusion?
Question 12 options:
-1.9035; conclude H1
-1.9422; conclude Ho
-1.9422; conclude H1
-1.9035; conclude Ho
To test the hypothesis, we can use the two-sample t-test for comparing the means of two populations. The null hypothesis (Ho) in this case would be that the mean cost of hotels in the downtown area of Vancouver is equal to the mean cost of hotels in Edmonton. The alternative hypothesis (H1) would be that the mean cost of hotels in Vancouver is less than the mean cost of hotels in Edmonton.
To find the standardized test statistic, we need to calculate the test statistic value using the sample means and standard deviations, and then compare it to the critical value for a given significance level.
The formula for the two-sample t-test is:
t = (mean1 - mean2) / sqrt((s1^2/n1) + (s2^2/n2))
Where:
- mean1 and mean2 are the sample means (Vancouver and Edmonton, respectively)
- s1 and s2 are the sample standard deviations (Vancouver and Edmonton, respectively)
- n1 and n2 are the sample sizes (Vancouver and Edmonton, respectively)
Given the following values:
- mean1 = $109, mean2 = $114
- s1 = $15, s2 = $12
- n1 = 50, n2 = 60
Substituting these values into the formula, we get:
t = (109 - 114) / sqrt((15^2/50) + (12^2/60))
= -5 / sqrt((225/50) + (144/60))
= -5 / sqrt(4.5 + 2.4)
= -5 / sqrt(6.9)
≈ -1.9035
The value of the standardized test statistic is approximately -1.9035.
To determine the correct conclusion, we need to compare this test statistic value to the critical value for a given significance level. Since the problem states a 5% level of significance, we need to find the critical value associated with a one-tailed test at the 5% level.
Looking up the critical value in a t-distribution table or using statistical software, we find that the critical value for a one-tailed test at the 5% level with degrees of freedom of (n1 + n2 - 2) = (50 + 60 - 2) = 108 is approximately -1.664.
Since the test statistic value (-1.9035) is less than the critical value (-1.664), we reject the null hypothesis (Ho) and conclude that there is evidence to support the claim that the mean cost of hotels in the downtown area of Vancouver is less than in Edmonton. Therefore, the correct conclusion is:
-1.9035; conclude H1