MATH SOLVE

9 months ago

Q:
# 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.)

Accepted Solution

A:

Answer:a. two-tailed testb. reject H0 if Z>1.96 or Z<-1.96c. Z = 1.73d. we have failed to reject the null hypothesise. P-value = 0.0418Step-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) = 0.025If 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<-1.96The 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 = 1.73Z<1.96, therefore we have failed to reject the null hypothesisThe 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