Question

Consider the hypothesis test H0:μ1=μ2 against H1:μ1<μ2. Suppose the sample sizes are n1=n2=15, that x1=6.2 and x2=7.8and that s21=4 and s22=6.25. Assume that σ21=σ22 and the data are drawn from normal distributions. Use α=0.05..

a) Test the hypothesis and find the p-value.
b) Explain how the test could be conducted with a confidence interval.
c)What is the power of the test in pair a) if μ1 is three unit less than μ2?

Answer 1

To test the hypothesis H0:μ1=μ2 against H1:μ1<μ2, we can use a t-test. The test statistic is calculated by dividing the difference in means by the standard error of the difference, and the p-value is found using the t-distribution table or a calculator. If the p-value is less than the significance level, we reject the null hypothesis.

Step-by-step explanation:

To test the hypothesis H0:μ1=μ2 against H1:μ1<μ2, we can use a t-test. Given that n1 = n2 = 15, x1 = 6.2, x2 = 7.8, s21 = 4, and s22 = 6.25, we can calculate the test statistic as:

t = (x1 - x2) / √((s21/n1) + (s22/n2))

Plugging in the values, we get t = (6.2 - 7.8) / √((4/15) + (6.25/15)) = -1.0865

The degrees of freedom for this test is (n1 + n2 - 2) = (15 + 15 - 2) = 28. Using a t-table or calculator, we can find the p-value for this test. The p-value is the probability of observing a t-value as extreme as -1.0865 or less. Comparing the p-value to the significance level α = 0.05, if the p-value is less than α, we reject H0. If the p-value is greater than or equal to α, we fail to reject H0. In this case, if the p-value is less than 0.05, we reject H0.