[The following information applies to the questions displayed below.] A sample o

Question

[The following information applies to the questions displayed below.] A sample of 49 observations is selected from a normal population. The sample mean is 61, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.10 significance level. H0: = 63 H1: 63 rev: 10_12_2016_QC_CS-65155 12.value: 1.00 pointsRequired information a. Is this a one- or two-tailed test? One-tailed test Two-tailed test ReferenceseBook & Resources Multiple ChoiceDifficulty: 2 IntermediateLearning Objective: 10-05 Conduct a test of a hypothesis about a population mean. Check my work 13.value: 1.00 pointsRequired information b. What is the decision rule? Reject H0 if -1.645 < z < 1.645 Reject H0 if z < -1.645 or z > 1.645 ReferenceseBook & Resources Multiple ChoiceDifficulty: 2 IntermediateLearning Objective: 10-05 Conduct a test of a hypothesis about a population mean. Check my work 14.value: 2.00 pointsRequired information c. What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) Value of the test statistic ReferenceseBook & Resources WorksheetDifficulty: 2 IntermediateLearning Objective: 10-05 Conduct a test of a hypothesis about a population mean. Check my work 15.value: 1.00 pointsRequired information d. What is your decision regarding H0? Fail to reject H0 Reject H0 ReferenceseBook & Resources Multiple ChoiceDifficulty: 2 IntermediateLearning Objective: 10-05 Conduct a test of a hypothesis about a population mean. Check my work 16.value: 2.00 pointsRequired information e. What is the p-value? (Round z value to 2 decimal places and final answer to 4 decimal places.) p-value

Explanation / Answer

Solution:

a. It is a two-tailed test because it is a non-directional hypothesis.

b. Reject H0 if z < -1.645 or z > 1.645

Using Z-tables, the critical value at a/2 = 0.10/2 = 0.05 is  ±1.645

c. Test Statistic

Z = (X-bar - µ)/ (/n)

Z = (61 – 63)/ (6/49)

Z = -2.33

d. Reject Ho

Because -2.33 < -1.645

e. P-value = 0.01

Using Z-tables, the p-value is

2 x P [Z< -2.33] = 2 x 0.0099 = 0.0198

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