Question

The null and alternate hypotheses are: H 0 ​ :μ 1 ​ ≤μ 2 ​ H 1 ​ :μ 1 ​ >μ 2 ​ A random sample of 22 items from the first population showed a mean of 109 and a standard deviation of 13. A sample of 18 items for the second population showed a mean of 101 and a standard deviation of 5 . Assume the sample populations do not have equal standard deviations and use the 0.01 significant level. Required: a. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) b. State the decision rule for 0.01 significance level. (Round your answer to 3 decimal places.) d. What is your decision regarding the null hypothesis? Do not reject Reject e. At the 0.01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? Yes No

Answer 1

To find the degrees of freedom for the unequal variance test, use the formula df = (s1^2 / n1 + s2^2 / n2)^2 / ((s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1)). The decision rule for the 0.01 significance level is to reject the null hypothesis if the test statistic is greater than the critical value. With the given data, the test statistic is approximately 2.450, which does not exceed the critical value.

Step-by-step explanation:

a. To find the degrees of freedom for an unequal variance test, you need to use the formula:

df = (s1^2 / n1 + s2^2 / n2)^2 / ((s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1))

Plugging in the values, we get:

df = (13^2 / 22 + 5^2 / 18)^2 / ((13^2 / 22)^2 / (22 - 1) + (5^2 / 18)^2 / (18 - 1))

Solving this equation, we find that the degrees of freedom is approximately 33.913, rounded down to the nearest whole number, the degrees of freedom is 33.

b. The decision rule for the 0.01 significance level is to reject the null hypothesis if the test statistic is greater than the critical value.

c. To determine the decision regarding the null hypothesis, we need to calculate the test statistic. Using the formula:

t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))

Plugging in the values, we get:

t = (109 - 101) / sqrt((13^2 / 22) + (5^2 / 18))

Solving this equation, we find that the test statistic is approximately 2.450.

Since the test statistic does not exceed the critical value, we fail to reject the null hypothesis.

e. At the 0.01 significance level, we do not have enough evidence to conclude that there is a difference in the mean number of times men and women order take-out dinners in a month.