Chapter 10: Hypothesis Testing: Additional Topics 1
Chapter 10:
Hypothesis Testing: Additional Topics
10.1n = 25 paired observations with sample means of 50 and 60 for populations 1 and 2. Can you reject the null hypothesis at an alpha of .05 if
a. = 20,
= 2.500, p-value is between .010 and .005.
Reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 10.0000 20.0000 4.0000
95% lower bound for mean difference: 3.1565
T-Test of mean difference = 0 (vs > 0): T-Value = 2.50 P-Value = 0.010
b. = 30,
= 1.67, p-value = .054. Do not reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 10.0000 30.0000 6.0000
95% lower bound for mean difference: -0.2653
T-Test of mean difference = 0 (vs > 0): T-Value = 1.67 P-Value = 0.054
c. = 15,
= 3.33, p-value = .001. Reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 10.0000 15.0000 3.0000
95% lower bound for mean difference: 4.8674
T-Test of mean difference = 0 (vs > 0): T-Value = 3.33 P-Value = 0.001
d. = 40,
= 1.25, p-value = .112. Do not reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 10.0000 40.0000 8.0000
95% lower bound for mean difference: -3.6871
T-Test of mean difference = 0 (vs > 0): T-Value = 1.25 P-Value = 0.112
10.2n = 25 paired observations with sample means of 50 and 56 for populations 1 and 2. Can you reject the null hypothesis at an alpha of .05 if
a. = 20,
= -1.50, p-value = .073. Do not reject at alpha of .05
Paired T-test and CI
N Mean StDev SE Mean
Difference 25 -6.00000 20.00000 4.00000
95% upper bound for mean difference: 0.84353
T-Test of mean difference = 0 (vs < 0): T-Value = -1.50 P-Value = 0.073
b. = 30,
= -1.00, p-value = .164. Do not reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 -6.00000 30.00000 6.00000
95% upper bound for mean difference: 4.26529
T-Test of mean difference = 0 (vs < 0): T-Value = -1.00 P-Value = 0.164
c. = 15,
= -2.00, p-value = .028. Reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 -6.00000 15.00000 3.00000
95% upper bound for mean difference: -0.86735
T-Test of mean difference = 0 (vs < 0): T-Value = -2.00 P-Value = 0.028
d. = 40,
= -.75, p-value = .230. Do not reject at alpha of .05
Paired T-Test and CI
N Mean StDev SE Mean
Difference 25 -6.00000 40.00000 8.00000
95% upper bound for mean difference: 7.68706
T-Test of mean difference = 0 (vs < 0): T-Value = -0.75 P-Value = 0.230
10.3
= 2.04, p-value = .043. Reject at alpha levels in excess of 4.3%
Paired T-Test and CI
N Mean StDev SE Mean
Difference 145 0.051800 0.305500 0.025370
95% CI for mean difference: (0.001654, 0.101946)
T-Test of mean difference = 0 (vs not = 0): T-Value = 2.04 P-Value = 0.043
10.4
= 2.239, p-value = .0301. Reject at levels in excess of 3%
Paired T-Test and CI: Male, Female
Paired T for Male - Female
N Mean StDev SE Mean
Male 8 46437.5 2680.1 947.5
Female 8 44962.5 2968.4 1049.5
Difference 8 1475.00 1862.99 658.66
95% lower bound for mean difference: 227.11
T-Test of mean difference = 0 (vs > 0): T-Value = 2.24 P-Value = 0.030
10.5 reject if t(5,.05) > 2.015
= 2.665. Reject at the 5% level
10.6
a. Reject if . For .
Do not reject at alpha of .05.
b. Reject if . For .
Reject at alpha of .05.
c. Reject if . For .
Do not reject at alpha of .05.
d. Reject if . For .
Do not reject at alpha of .05.
10.7
a. Reject if . For the given data, .
Do not reject at alpha of .05.
b. Reject if . For the given data, .
Do not reject at alpha of .05.
c. Reject if . For the given data, .
Do not reject at alpha of .05.
d. Reject if . For the given data, .
Do not reject at alpha of .05.
10.8
= 7.334.
Reject at all common levels of alpha
10.9
= 12.96.
Reject at all common levels of alpha
10.10
= -1.0207,
p-value = 2[1-FZ(1.02)] = 2[1-.8461] = .3078
Therefore, reject at levels of alpha in excess of 30.78%
10.11
= = 642.66925
= = -1.8995
(1.8995)=.0308; p-value = 2(.0308) = .0616.
Reject at levels in excess of 6.16%
10.12Assuming both populations are normal with equal variances:
= 29.592247
= = 1.108
Therefore, do not reject at the 10% alpha level since 1.108 < 1.645 = t(119,.05)
10.13
= 3,632,605, = 1.275
Therefore, do not reject at the 10% alpha level since 1.275 < 1.33 = t(18,.1)
10.14 a.
= .4636, = -2.65 p-value = .004. Therefore, reject at all common levels of alpha
b.
= .6218,
= -1.36
p-value = .0869. Therefore, reject at .10, but do not reject at the .05 level
c.
= .4582,
= -2.32
p-value = .0102.
Therefore, reject at the .05 level, but do not reject at the .01 level
d.
= .299, = -3.25
p-value = .0006. Therefore, reject at all common levels of alpha
e.
= .4064,
= -1.01
p-value = .1562. Therefore, do not reject at any common level of alpha
10.15
= .63,
= -2.63
p-value = .0043. Therefore, reject at all common levels of alpha
10.16
= .44, = -6.97
Reject at all common levels of alpha
10.17 reject if |z.025| > 1.96
= .266
= -1.466. Do not reject at the 5% level
10.18 reject if |z.025| > 1.96
= .36714
= 2.465. Reject at the 5% level
10.19
= .614
= -8.216.
Reject at all common levels of alpha
10.20 reject if |z.05| > 1.645
= .554
= .926.
Do not reject at the 5% level
10.21 reject if z.01 < -2.33
= .577
= -18.44. Reject at the 1% level
10.22 a.
F = 125/51 = 2.451. Reject at the 1% level since 2.451 > 2.11 F(44,40,.01)
b.
F = 235/125 = 1.88. Reject at the 5% level since 1.88 > 1.69 F(43,44,.05)
c.
F = 134/51 = 2.627. Reject at the 1% level since 2.627 > 2.11 F(47,40,.01)
d.
F = 167/88 = 1.90. Reject at the 5% level since 1.90 > 1.79 F(24,38,.05)
10.23
F = 1614.208/451.770 = 3.573.
Reject at the 1% level since 3.573 > 2.41F(29,29,.01)
10.24; reject if F(3,6,.05) > 4.76
F = 114.09/16.08 = 7.095. Reject at the 5% level
10.25;
F=(27.56)2/(22.93)2=1.44.
Do not reject at the 10% level since 1.44<1.84F(35,35,.05)
10.26;
F = (2107)2/(1681)2 = 1.57
Therefore, do not reject at the 10% level since 1.57 < 3.18 F(9,9,.05)
10.27;
F = (24.4)2/(20.2)2 = 1.46.
Do not reject at the 5% level since 1.46 < 9.28 F(3,3,.05)
10.28 No. The probability of rejecting the null hypothesis given that it is true is 5%.
10.29
= = 3.4525
= = 1.966
p-value is between (.025, .010) x 2 = .05 and .02.
Reject at levels in excess of 5%
10.30a. reject if t.05 > 1.671
= 2.574. Reject at the 5% level
b. reject if t .05 < -1.645
= = 1.853
= = -4.293.
Reject at levels in excess of 5%
10.31 reject if |t .05| > 1.645
= = 347.980
= =-.210.
Do not reject at levels in excess of 5%
10.32Presuming the populations are normally distributed with equal variances, the samples must be independent random samples:
reject if t(6,.01) < -3.143
= = 106.58
= = -5.027.
Reject at levels in excess of 1%
10.33Assuming the populations are normally distributed with equal variances and independent random samples:
Magazine A: , Magazine B:
reject if t(10,.05) > 1.812
= = 4.8416
= = 3.330.
Reject at levels in excess of 5%
10.34Assume that the populations are normally distributed with equal variances and independent random samples:
Magazine A: , Magazine B:
= = 6.4416
= = .183. Do not reject at any common level of alpha
10.35 Sample sizes greater than 100, use the z-test.
= -2.30, p-value = 1 – FZ(2.3) = 1 - .9893 = .0107
Therefore, reject at levels of alpha in excess of 1.07%
10.36 . Sample sizes less than 100, use the t-test
= = .32675
= = -1.901. p-value is between (.05 and .025) x 2 = .10 and .05. Reject at any alpha of .10 or higher.
10.37a. reject if z.05 < -1.645
= -1.2. Do not reject at the 5% level
b. reject Ho if |z.025| > 1.96
= .478, = .932
Therefore, do not reject at the 5% level
10.38 reject Ho if t(44,.05) < -1.684
= = .00319
= = -5.284.
Reject at any common level of alpha
10.39 reject Ho if |z.01| < -2.33
=.211, = -1.19.
Do not reject at the 1% level
10.40
= .630435, = 1.235, p-value =
2[1-FZ(1.24)] = 2[1-.8925] = .1075.
Reject at levels of alpha in excess of 10.75%
10.41
= .617, = -1.653,
p-value = 1–FZ(1.65)]=.0495
Therefore, reject at levels of alpha in excess of 4.95%
10.42; , = (2.647)2/(1.63656)2 = 2.618. Do not reject at the 5% level, 2.618 < 5.05 F(5,5,.05)
10.43; , = (4.16314)2/(4.05421)2 = 1.0545. Do not reject at the 5% level, 1.0545 < 2.98 F(10,10,.05). There is insufficient evidence to suggest that the population variances differ between the two forecasting analysts.
10.44 a.
df = n1 + n2 – 2 = 27 + 27 – 2 = 52; t52,.05 = 1.675
At the .05 level of significance, reject Ho and accept the alternative that the mean output per hectare is significantly greater with the new procedure.
- 95% acceptance interval:
, , , because F calc is within the acceptance interval, there is not sufficient evidence against the null hypothesis that the sample variances are not significantly different from each other.
10.45 a. reject if |z.015| > 2.17
= .3453, = 1.987
Therefore, reject at the 5% level, but do not reject at the 3% level
b. reject if |z.03| > 1.88
= .3453, = 1.987
Therefore, reject at the 3% level
10.46 Assume that the population of matched differences are normally distributed
reject if |t(9,.05)| > 1.833
of the matched differences = 1.13, s of the matched differences = 1.612
= 2.22, p-value =.054.
Reject at the 10%, but not the 5% level
Paired T-Test and CI: VARIETY A, VARIETY B
Paired T for VARIETY A - VARIETY B
N Mean StDev SE Mean
VARIETY A 10 11.9300 2.9265 0.9254
VARIETY B 10 10.8000 2.5237 0.7981
Difference 10 1.13000 1.61180 0.50969
95% CI for mean difference: (-0.02301, 2.28301)
T-Test of mean difference = 0 (vs not = 0): T-Value = 2.22
P-Value = 0.054
10.47a. The box plots of the raw data show similar medians and interquartile ranges for both brands. However, brand 2 is dominated by three outliers that are skewing the brand 2 data to the right:
The descriptive statistics show the effect of the extreme outliers on brand 2 sales —note the sizeable standard deviation of brand 2:
Descriptive Statistics: saleb2, saleb4
Variable N Mean Median TrMean StDev SE Mean
saleb2 52 181.2 127.0 155.7 154.9 21.5
saleb4 52 140.29 125.50 136.80 60.84 8.44
Variable Minimum Maximum Q1 Q3
saleb2 59.0 971.0 94.8 203.3
saleb4 55.00 305.00 101.25 182.75
The matched pairs t-test on the original data shows a significant difference between the weekly sales with brand 2 found to be significantly larger than brand 4 at the .05 level:
Paired T-Test and CI: saleb2, saleb4
Paired T for saleb2 - saleb4
Variable N Mean StDev SE Mean
saleb2 52 181.2 154.9 21.5
saleb4 52 140.3 60.8 8.4
Difference 52 40.9 169.5 23.5
95% lower bound for mean difference: 1.5
T-Test of mean difference = 0 (vs > 0): T-Value = 1.74 P-Value = 0.044
- However, with only the largest outlier removed from the data of brand 2, the difference between the two brands becomes insignificant at the .05 level:
Paired T-Test and CI: saleb2_1, saleb4 (with outlier removed)
Paired T for saleb2_1 - saleb4
N Mean StDev SE Mean
saleb2_1 51 165.7 108.5 15.2
saleb4 51 140.8 61.3 8.6
Difference 51 24.9 125.7 17.6
95% lower bound for mean difference: -4.6
T-Test of mean difference = 0 (vs > 0): T-Value = 1.42 P-Value = 0.081
10.48 a.
Results for: Ole.MTW
Two-Sample T-Test and CI: Olesales, Carlsale
Two-sample T for Olesales vs Carlsale
N Mean StDev SE Mean
Olesales 156 3791 5364 429
Carlsale 156 2412 4249 340
Difference = mu Olesales - mu Carlsale
Estimate for difference: 1379
95% lower bound for difference: 475
T-Test of difference = 0 (vs >): T-Value = 2.52 P-Value = 0.006 DF = 310
Both use Pooled StDev = 4839
Reject H0 at the .01 level of significance.
b.
Two-Sample T-Test and CI: Oleprice, Carlpric
Two-sample T for Oleprice vs Carlpric
N Mean StDev SE Mean
Oleprice 156 0.819 0.139 0.011
Carlpric 156 0.819 0.120 0.0096
Difference = mu Oleprice - mu Carlpric
Estimate for difference: -0.0007
95% CI for difference: (-0.0297, 0.0283)
T-Test of difference = 0 (vs not =): T-Value = -0.05 P-Value = 0.962 DF = 310
Both use Pooled StDev = 0.130
Do not reject H0 at any common level of significance. Note that the 95% confidence interval contains 0, therefore, no evidence of a difference.
10.49The equation for an acceptance interval is shown next:
Since the package weights are not independent (), the variance of the sample means is given by the followingequation:
Calculate the variance of the sample means using and Also, use and
Thus, the standard deviation of the sample means is
For a 99% acceptance interval, so
The 99% acceptance interval is or (15.52, 16.48). The acceptance interval can be used for quality control monitoring of the process. The interval is plotted over time and provides limits for the sample mean
10.50 reject if or
Let and Then, and
Do not reject at the 5% level. Conclude that there is a not a difference in the proportion of humorous ads in British versus American trade magazines.