# A sample of 38 observations is selected from one population with a population standard deviation of 4.9. The sample mean is 102.0. A sample of 49 observations is selected from a second population with a population standard deviation of 4.8. The sample mean is 100.4. Conduct the following test of hypothesis using the 0.04 significance level. H0 : μ1 = μ2H1 : μ1 ≠ μ2a) State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) The decision rule is to reject Ho is z is________ the interval (__________, _____)b) Compute the value of the test statistic. (Round your answer to 2 decimal places.)c) What is the p-value? (Round your answer to 4 decimal places.)

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A sample of 38 observations is selected from one population with a population standard deviation of 4.9. The sample mean is 102.0. A sample of 49 observations is selected from a second population with a population standard deviation of 4.8. The sample mean is 100.4. Conduct the following test of hypothesis using the 0.04 significance level.

H0 : μ1 = μ2

H1 : μ1μ2

a) State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) The decision rule is to reject Ho is z is________ the interval (__________, _____)

b) Compute the value of the test statistic. (Round your answer to 2 decimal places.)

c) What is the p-value? (Round your answer to 4 decimal places.)

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Step 1

The sample means of first and second population are 102.3 and 100.4 respectively. The standard deviations of two populations are 4.9 and 4.8 respectively. The sample sizes of first and second populations are 38 and 49 respectively.

Step 2

The significance level is 0.04.

Since the hypothesis test is two tailed test. The critical value at the significance 0.04 can be obtained using Excel formula = NORM.S.INV (0.02)

Step 3

a).

Decision rule:

Reject the null hypothesis at the ...

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