Chapter 9
CHAPTER 9 – HYPOTHESIS TESTING AND ESTIMATION FOR TWO POPULATION VALUES
1.
a. If the calculated F > 4.405, reject Ho, otherwise do not reject Ho
b. F = 232/192 = 1.4654
Since 1.4654 < 4.405 do not reject Ho
2.
a. If the calculated F > 2.255, reject Ho, otherwise do not reject Ho
b. F = 2302/2102 = 1.1995
Since 1.1995 < 2.255 do not reject Ho
3. s1 = 2.8975 s2 = 2.7033
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 3.179, reject Ho, otherwise do not reject Ho
F = 2.89752/2.70332 = 1.1488
Since 1.1488 < 3.179 do not reject Ho
4.
a. H0: σd2 = σw2
HA: σd2 ≠ σw2
b. If the calculated F > 3.284, reject Ho, otherwise do not reject Ho
F = 22/1.22 = 2.7778
Since 2.7778 < 3.284 do not reject Ho and conclude that there is no difference in the standard deviations
5.
a. H0: σm2 σf2
HA: σm2 > σf2
b. If the calculated F > 1.984, reject Ho, otherwise do not reject Ho
F = 2.52/1.342 = 3.4807
Since 3.4807 > 1.984 reject Ho and conclude that there is more variability in male donations than in female donations.
6. sA = 7.1375 sB = 8.6929
H0: σA2 = σB2
HA: σA2 ≠ σB2
If the calculated F > 3.0274, reject Ho, otherwise do not reject Ho
F = 8.69292/7.13752 = 1.4833
Since 1.4833 < 3.0274 do not reject Ho and conclude that there is no difference in the standard deviation of dollars returned between the two brochures.
7.
Sales Plan / Data / TotalBasic Plan / Count of Cell Phone Minutes / 52.000
StdDev of Cell Phone Minutes / 34.434
Business Plan / Count of Cell Phone Minutes / 148.0000
StdDev of Cell Phone Minutes / 29.9187
a. H0: σBasic2 σBus.2
HA: σBasic2 > σBus2
b. If the calculated F > 1.4341, reject Ho, otherwise do not reject Ho
F = 34.4342/29.91872 = 1.3246
Since 1.3246 < 1.4341 do not reject Ho and conclude that the standard deviation in minutes used by the Business Plan is not less than the Basic Plan
8. df = 100 + 120 – 2 = 218
a. If t > 1.96 or t < -1.96 reject Ho, otherwise do not reject Ho
b. Since the degrees of freedom are not available in the F table you must use Excel’s FINV function to find the critical F. If the calculated F > 1.4658, reject Ho, otherwise do not reject Ho
F = 242/202 = 1.44
Since 1.44 < 1.4658 do not reject Ho and conclude that the variances are equal
c. sp = = 22.2727
t = (430 – 405)/(22.2727) = 8.2898
Since 8.2898 > 1.96 reject Ho
9.
a. If the calculated F > 2.108, reject Ho, otherwise do not reject Ho
b. H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 2.108 (using α = 0.10, critical value for α = 0.05 not available in tables), reject Ho, otherwise do not reject Ho
F = 322/302 = 1.1378
Since 1.1378 < 2.108 do not reject Ho and conclude that the variances are equal
10.
a. Ho: μd 0
Ha: μd > 0
b.
Sample 1 / Sample 2 / Difference (d) / (d - )250 / 38 / 12 / 36
47 / 44 / 3 / 9
44 / 38 / 6 / 0
48 / 37 / 11 / 25
40 / 43 / -3 / 81
36 / 44 / -8 / 196
43 / 31 / 12 / 36
46 / 38 / 8 / 4
72 / 39 / 33 / 729
40 / 54 / -14 / 400
55 / 41 / 14 / 64
38 / 40 / -2 / 64
72 / 1644
= 72/12 = 6
sd = = 12.2252
t = (6 – 0)/(12.2252/) = 1.7001
Decision Rule:
If t > 2.7181 reject Ho otherwise do not reject Ho
Since t = 1.7001 < 2.7181 do not reject Ho and conclude that the mean of Population 1 is not greater than the mean of Population 2
c. Must first test for equal variances at alpha = 0.02 (must use Excel’s FINV function)
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 4.462, reject Ho, otherwise do not reject Ho
F = 9.64332/5.53432 = 3.0362
Since 3.3062 < 4.462, fail to reject Ho and conclude equal variances so can use pooled standard deviation
Ho: μ1 – μ2 0
Ha: μ1 – μ2 > 0
df = 12 + 12 – 2 = 22
Using Excel’s average and stdev functions students can determine the sample mean and sample standard deviation of each of the samples
Sample 1: Sample 2:
Mean = 46.5833 Mean = 40.5833
St. Dev. = 9.6433 St. Dev. = 5.5343
If t > 2.5083 reject Ho, otherwise do not reject Ho
sp = = 7.8620
t = (46.5833 – 40.5833)/(7.8620) = 1.8694
Since 1.8694 < 2.5083 do not reject Ho and conclude that the mean of Population 1 is not greater than the mean of Population 2; same conclusion as reached in part b.
11.
a. Ho: μd = 0
Ha: μd ≠ 0
b.
Sample 1 / Sample 2 / Difference (d) / (d - )24.4 / 3.7 / 0.7 / 4.6464
2.7 / 3.5 / -0.8 / 0.4298
1.0 / 4.0 / -3.0 / 2.3853
3.5 / 4.9 / -1.4 / 0.0031
2.8 / 3.1 / -0.3 / 1.3353
2.6 / 4.2 / -1.6 / 0.0209
2.4 / 5.2 / -2.8 / 1.8075
2.0 / 4.4 / -2.4 / 0.8920
2.8 / 4.3 / -1.5 / 0.0020
-13.1 / 11.5222
= -13.1/9 = -1.4556
sd = = 1.2001
t = (-1.4556 – 0)/(1.2001/) = -3.6387
Decision Rule:
If t > 1.8595 or t < -1.8595 reject Ho otherwise do not reject Ho
Since t = -3.6387 < -1.8595 reject Ho and conclude that the means of the populations are different
c. –1.4556 + 1.8595(1.2001/); -2.1995 ------0.7117; yes this is consistent with the answer in part b because the interval range does not include 0 which would mean that the mean of the differences is significantly different from 0.
12. You must first determine whether the population standard deviations are assumed to be equal. If tested at a significance level of 0.02 the results are as follows:
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 2.6591, reject Ho, otherwise do not reject Ho
F = 0.082/0.062 = 1.7778
Since 1.7778 < 2.6591 do not reject Ho and conclude that the standard deviations are equal
sp = = 0.0707
a. (0.145 – 0.107) + 1.6772(0.0707) ; 0.0045 ----- 0.0715; because 0 is not included in the confidence interval you would conclude that the means are different and since you are in the positive range you can conclude that the mean of Population 1 is greater than the mean of Population 2
b. (0.145 – 0.107) + 2.0106(0.0707) ; -0.0022 ----- 0.0782; because 0 is in this range you would conclude that the means are the same.
c. sp = = 0.0707
90% confidence interval:
(0.145 – 0.107) + 1.6606(0.0707) ; 0.0145 ----- 0.0615
95% confidence interval:
(0.145 – 0.107) + 1.9845(0.0707) ; 0.0099 ----- 0.0661
The pooled standard deviation does not change because it is essentially a weighted average of the two standard deviations and since you doubled both sample sizes the weights were not changed. The critical t-value changes because the sample sizes change and the reciprocals of the sample sizes change.
13.
a. Ho: μF – μM 1
Ha: μF – μM > 1
df = 60 + 60 – 2 = 118
If t > 1.6579 reject Ho, otherwise do not reject Ho
Must first determine if population standard deviations are equal. Test at alpha = 0.02
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 1.8459, reject Ho, otherwise do not reject Ho
F = 1.56 2/1.22 = 1.69
Since 1.69 < 1.8459 do not reject Ho and conclude that the standard deviations are equal
sp = = 1.3917
t = ((14.65 – 13.24) – 1)/(1.3917) = 1.6136
Since 1.6136 < 1.6579 do not reject Ho and conclude that the difference is not greater than 1
b. P(t > 1.6136) = 0.0546; p-value
14.
a. Ho: μC – μR 0.25
Ha: μC – μR > 0.25
df = 25 + 25 – 2 = 48
If t > 1.6772 reject Ho, otherwise do not reject Ho
Must first determine if population standard deviations are equal. Test at alpha = 0.02
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 2.6591, reject Ho, otherwise do not reject Ho
F = 0.87 2/0.792 = 1.2128
Since 1.2128 < 2.6591 do not reject Ho and conclude that the standard deviations are equal
sp = = 0.8310
t = ((3.74 – 3.26) – 0.25)/(0.8310) = 0.9785
Since 0.9785 < 1.6772 do not reject Ho and conclude that the difference is not greater than $0.25
b. Since you accepted the null hypothesis the type of error that could occur is accepting a false null hypothesis that is a Type II error.
15.
a. Ho: μN – μO 0
Ha: μN – μO > 0
df = 35 + 30 – 2 = 63
If t > 1.2951 reject Ho, otherwise do not reject Ho
Must first determine if population standard deviations are equal. Test at alpha = 0.02
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 2.3716, reject Ho, otherwise do not reject Ho
F = 16.23 2/15.912 = 1.0406
Since 1.0406 < 2.3716 do not reject Ho and conclude that the standard deviations are equal
sp = = 16.0835
t = ((288 – 279) – 0)/(16.0835) = 2.2491
Since 2.2491 > 1.2951 reject Ho and conclude that the new cartridge will result in a longer lasting product.
b. 90% confidence interval:
(288 – 279) + 1.6694(16.0835) ; 2.3196 ----- 15.6804; yes this is consistent with the results in part a.
16.
Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 1.351, reject Ho, otherwise do not reject Ho
F = 6.18 2/2.332 = 7.035
Since 7.035 > 1.351 reject Ho and conclude that the standard deviations are not equal
a. 90% confidence interval:
(22.48 – 12.56) + 1.645; 9.2012 ----- 10.6388
95% confidence interval:
(22.48 – 12.56) + 1.96; 9.0635 ----- 10.7765
b. Yes, since both confidence intervals contain the value of $10 this means the fee could be $10 more for federally chartered banks.
17. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 1.3184, reject Ho, otherwise do not reject Ho
F = 32 2/112 = 8.4628
Since 8.4628 > 1.3184 reject Ho and conclude that the standard deviations are not equal
a.
90% confidence interval:
(53 – 24) + 1.645; 25.9624 ----- 32.0376; yes there is a difference in how long males and females spend in the store per visit because the confidence interval does not contain the value 0.
b. (53 – 24) + 1.96; 25.3808 ----- 32.6192; no because the difference was so large in part a that changing the confidence interval to 95% would not change the upper and lower limits enough to cause them to include the value 0
c. Based upon the confidence intervals calculated in parts a and b it would actually be more than 20 minutes more.
d. P(men > 53) = P(z > (53 – 24)/(11/)) = P(z > 41.68) which is essentially 0
18. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 1.3923, reject Ho, otherwise do not reject Ho
F = 5 2/3.62 = 1.929
Since 1.929 > 1.3923 reject Ho and conclude that the standard deviations are not equal
a.
(41.5 – 39) + 1.96; 1.6461 ----- 3.3539; yes because the interval does not contain the value 0 which would indicated no difference.
b. Company A:
-1.645 = (x – 41.5)/(3.6/); x = 41.0813
Company B:
-1.645 = (x – 39)/(5/); x = 38.4184
19. Must first determine if population standard deviations are equal. Test at alpha = 0.02. Because of the sample sizes you must use Excel’s FINV function.
H0: σ12 = σ22
HA: σ12 ≠ σ22
If the calculated F > 1.9626, reject Ho, otherwise do not reject Ho
F = 1.05 2/0.892 = 1.392
Since 1.392 < 1.9626 do not reject Ho and conclude that the standard deviations are equal
a. 5.26 – 6.19 = -0.93; The advantage of using this point estimate is that it is easy to calculate. The disadvantage is that it uses only 2 values calculated from the samples and it does not consider the spread or the variation in the sample at all.