Question

A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.67. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59. Conduct the following test of hypothesis using the 0.08 significance level.

H0 : μ1 ≤ μ2
H1 : μ1 > μ2

a. Is this a one-tailed or a two-tailed test?
b. State the decision rule.
c. Compute the value of the test statistic.
d. What is your decision regarding H0?
e. What is the p-value?

Answer 1

This is a one-tailed test with a significance level of 0.08. The null hypothesis is rejected based on the test statistic and the decision rule. The p-value represents the probability of observing a difference in means as extreme as the observed difference, assuming the null hypothesis is true.

Step-by-step explanation:

a. This is a one-tailed test because the alternative hypothesis specifies a direction (μ1 > μ2).

b. The decision rule depends on the significance level. Since the significance level is 0.08, we compare the p-value to 0.08. If the p-value is less than 0.08, we reject the null hypothesis.

c. To compute the test statistic, we first calculate the standard error of the difference in means using the formula: SE = sqrt((s1^2/n1) + (s2^2/n2)), where s1 and s2 are the sample standard deviations and n1 and n2 are the sample sizes. Then, we calculate the test statistic as: test_statistic = (sample_mean1 - sample_mean2) / SE.

d. Based on the test statistic and the decision rule, we reject the null hypothesis (H0 : μ1 ≤ μ2).

e. The p-value is the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true. We can find the p-value by comparing the test statistic to a t-distribution with degrees of freedom equal to the sum of the sample sizes minus 2.