Question 191029: Did I do this right?
Hypothesis Testing for the Mean (Large Samples)
Convicted murderers receive a sentence of an average of 18.7 years in prison. A criminologist wants to perform a hypothesis test to determine whether the mean sentence by one particular judge differs from 18.7 years. A random sample of 40 cases from the court files from this judge is taken. It is found that sample mean is 17.2 years. Assume that the population standard deviation is 7.4 years. Test at the 0.05 significance level.
Solution: Criminologist wants to perform a hypothesis test to determine whether the mean sentence by one particular judge differs from 18.7 years
1. H0: µ = 18.7
Ha: µ ≠18.7
2. significance level = 0.05
3. Test statistics: The test is two-tailed
z = xbar - µ / ơ/√n = (17.2-18.7)/[7.4/sqrt(40)] = -1.2820
4. P-value or critical z0 or t0. The critical values are – z0 = -1.96 and z0 = 1.96
5. Rejection Region: z < -1.96 and z > 1.96
6. Decision: since test statistic is not in the reject interval, Fail to reject Ho.
7. Interpretation: The test does not provide strong evidence for rejecting the belief the average sentence is 18.7
b. Use the P-value method.
1. H0: µ = 18.7
Ha: µ ≠18.7
2. significance level = 0.05
3. Test statistics: z = xbar - µ / ơ/√n = (17.2-18.7)/[7.4/sqrt(40)] = -1.2820
4. P-value or critical z0 or t0. In Table 4, the area corresponding to z = -1.2820 is 0.1003. Because the test is a two-tailed test, the P-value is equal to twice the area to the left of z = 0.2820
P = 2(0.1003) = 0.2006
5. Rejection Region: ?
6. Decision: Since p-value is greater than 0.05 fail to reject Ho.
7. Interpretation: The test does not provide strong evidence for rejecting the belief the average sentence is 18.7
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Did I do this right?
Hypothesis Testing for the Mean (Large Samples)
Convicted murderers receive a sentence of an average of 18.7 years in prison. A criminologist wants to perform a hypothesis test to determine whether the mean sentence by one particular judge differs from 18.7 years. A random sample of 40 cases from the court files from this judge is taken. It is found that sample mean is 17.2 years. Assume that the population standard deviation is 7.4 years. Test at the 0.05 significance level.
Solution: Criminologist wants to perform a hypothesis test to determine whether the mean sentence by one particular judge differs from 18.7 years
1. H0: µ = 18.7
Ha: µ ≠18.7
2. significance level = 0.05
3. Test statistics: The test is two-tailed
z = xbar - µ / ơ/√n = (17.2-18.7)/[7.4/sqrt(40)] = -1.2820
4. P-value or critical z0 or t0. The critical values are – z0 = -1.96 and z0 = 1.96
5. Rejection Region: z < -1.96 and z > 1.96
6. Decision: since test statistic is not in the reject interval, Fail to reject Ho.
7. Interpretation: The test does not provide strong evidence for rejecting the belief the average sentence is 18.7
b. Use the P-value method.
1. H0: µ = 18.7
Ha: µ ≠18.7
2. significance level = 0.05
3. Test statistics: z = xbar - µ / ơ/√n = (17.2-18.7)/[7.4/sqrt(40)] = -1.2820
4. P-value or critical z0 or t0. In Table 4, the area corresponding to z = -1.2820 is 0.1003. Because the test is a two-tailed test, the P-value is equal to twice the area to the left of z = 0.2820
P = 2(0.1003) = 0.2006
5. Rejection Region: (-10^9 < t < -1.2820)
6. Decision: Since p-value is greater than 0.05 fail to reject Ho.
7. Interpretation: The test does not provide strong evidence for rejecting the belief the average sentence is 18.7
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Your work looks good to me although the t-distribution is generally
recommended these days for tests of means. It looks like your text
required you to use a z-test-----which is alright.
Cheers,
Stan H.
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