Question 191043: I started most of the work but I am not sure if I am doing it right. Please help.
Hypothesis Testing for Mean (Small Samples)
3. Metro Bank claims that the mean wait time for a teller during peak hours is less than 4 minutes. A random sample of 20 wait times has a mean of 2.6 minutes with a sample standard deviation of 2.1 minutes.
a. Use the critical value z0 method from the normal distribution to test for the population mean . Test the company’s claim at the level of significance = 0.05.
xbar = 2.6
µ = 4
s = 2.1
n = 20
d.f = n – 1 = 20 – 1 = 19
To find the critical value, use table 5 in Appendix B with d.f. = 19 and 0.05 in the “One Tail, ” column. Because the test is a left-tailed test, the critical value is negative. So
t0 = -1.729
1. H0 : u >= 4 minutes
Ha : u < 4 minutes
2. level of significance = 0.05
3. Test statistics: t = xbar - µ / s/√n (2.6-4)/[2.1/sqrt(20)] = -2.9814
4. P-value or critical z0 or t0.
5. Rejection Region: t < -1.729
6. Decision: Since -2.9814 is in the reject interval, Reject Ho.
7. Interpretation: The mean time is not <= 4 minutes
b. Use the critical value z0 method from the normal distribution to test for the population mean. Test the company’s claim at the level of significance = 0.01.
xbar = 2.6
µ = 4
s = 2.1
n = 20
d.f = n – 1 = 20 – 1 = 19
To find the critical value, use table 5 in Appendix B with d.f. = 19 and 0.01 in the “One Tail, level of significance ” column. Because the test is a left-tailed test, the critical value is negative. So
t0 = 2.539
1. H0 : u >= 4 minutes
2. Ha : u < 4 minutes
3. level of significance = 0.01
4. Test statistics:
5. P-value or critical z0 or t0.
6. Rejection Region:
7. Decision:
8. Interpretation:
Hypothesis Testing for Proportions.
4. In a recent poll, it was found that 43% of registered U.S. voters would vote for the incumbent president. If 100 registered voters were sampled randomly, it was found that 35% would vote of the incumbent. Test the claim that the actual proportion is 43%.
1. H0 :
Ha :
2. level of significance =
3. Test statistics:
4. P-value or critical z0 or t0.
5. Rejection Region:
6. Decision:
7. Interpretation:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 3. Metro Bank claims that the mean wait time for a teller during peak hours is less than 4 minutes. A random sample of 20 wait times has a mean of 2.6 minutes with a sample standard deviation of 2.1 minutes.
a. Use the critical value z0 method from the normal distribution to test for the population mean . Test the company’s claim at the level of significance = 0.05.
xbar = 2.6
µ = 4
s = 2.1
n = 20
d.f = n – 1 = 20 – 1 = 19
To find the critical value, use table 5 in Appendix B with d.f. = 19 and 0.05 in the “One Tail, ” column. Because the test is a left-tailed test, the critical value is negative. So
t0 = -1.729
1. H0 : u >= 4 minutes
Ha : u < 4 minutes
2. level of significance = 0.05
3. Test statistics: t = xbar - µ / s/√n (2.6-4)/[2.1/sqrt(20)] = -2.9814
4. P-value or critical z0 or t0.
5. Rejection Region: t < -1.729
6. Decision: Since -2.9814 is in the reject interval, Reject Ho.
7. Interpretation: The mean time is not <= 4 minutes
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part "a" looks good
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b. Use the critical value z0 method from the normal distribution to test for the population mean. Test the company’s claim at the level of significance = 0.01.
xbar = 2.6
µ = 4
s = 2.1
n = 20
d.f = n – 1 = 20 – 1 = 19
To find the critical value, use table 5 in Appendix B with d.f. = 19 and 0.01 in the “One Tail, level of significance ” column. Because the test is a left-tailed test, the critical value is negative. So
t0 = 2.539
1. H0 : u >= 4 minutes
2. Ha : u < 4 minutes
3. level of significance = 0.01
4. Test statistics:
5. P-value or critical z0 or t0.
6. Rejection Region:
7. Decision:
8. Interpretation:
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Hypothesis Testing for Proportions.
4. In a recent poll, it was found that 43% of registered U.S. voters would vote for the incumbent president. If 100 registered voters were sampled randomly, it was found that 35% would vote of the incumbent. Test the claim that the actual proportion is 43%.
1. H0 : p = 0.43
Ha : p is not 0.43
--------------------------
2. level of significance = alpha = 5%,
3. Test statistics: z(0.35) = (0.35-0.43)*sqrt[0.43*0.57/100] = -1.6159
4. P-value or critical z0 or t0.l: 2P(z<-1.6159) = 0.10611
5. Rejection Region: z<1.96 or z>1.96
6. Decision: Since the p-value is greater than 5%, Fail to reject Ho.
7. Interpretation: The test does not provide evidence that
lead to rejecting the poll results.
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Cheers,
Stan H.
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