Question

A sample of 38 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.5. A sample of 51 observations is selected from a second population with a population standard deviation of 5.8. The sample mean is 98.8. Conduct the following test of hypothesis using the 0.05 significance level. H0 : μ1 = μ2 H1 : μ1 ≠ μ2
a. This is a -tailed test.
b. State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)
c. Compute the value of the test statistic. (Round your answer to 2 decimal places.
d. What is your decision regarding H0?
e. What is the p-value? (Round your answer to 4 decimal places.)

Answer 1

Answer:

a. two-tailed test

b. reject H0 if Z>1.96 or Z<-1.96

c. Z = 1.73

d. we have failed to reject the null hypothesis

e. P-value = 0.0418

Step-by-step explanation:

We will use a Z test to resolve this, an it will be a two-tailed test because the hypothesis statements are not indicating a specific direction for the significant difference (H0 : μ1 = μ2 ; H1 : μ1 ≠ μ2), this also means that the significanclevel will be divided between the both tails (2.5% en each tail for the rejection regions). See attached drawing for reference.

We need to find our critical value:

Zα/2 = Z(0.05/2) = 0.025

If we look for 0.025 in a Z table we will find that the critical value is 1.96 to the right, and by symmetry -1.96 to the left. So our decision rule will be to reject H0 if Z>1.96 or Z<-1.96

The Z test will be done using the next equation:

Z = (x⁻1 - x⁻2) - (μ1 - μ2) / √( σ²1/n1 + σ²2/n2

Because we are testing the null hypothesis we know that μ1 - μ2 must be zero if they are supposed to be equal (H0 : μ1 = μ2), so we calculate as follows:

Z = (100.5 - 98.8) - (0) / √( (3.4)²/38 + (5.8)²/51 = 1.7/0.9817 = 1.73

Z<1.96, therefore we have failed to reject the null hypothesis

The P-value for this test would be represented by the probability of Z being greater than 1.73, so we can look for it in any Z table, finding that its value is 0.0418, and because P-value>0.025 we again confirm that we don't have evidence statistically significant to reject the null hypothesis