Chapter 6: Statistical Inference
1.
The population mean is a fixed number not a random variable, therefore it is incorrect to state that it has a
probability to lie in some range of values. It is correct to say that you are “95% confident that the
population mean lies between -5 and 5” since this refers to the method by which you are estimating the
confidence interval.
2.
False. Accepting the null hypothesis does not mean that the null hypothesis is true. It means that there is
insuffient evidence to reject it, accepting the alternative hypothesis.
3.
False. Rejecting the null hypothesis means that the probability of error associated with incorrectly
rejecting the null hypothesis is low enough to allow you to accept the alternative hypothesis. However, the
null hypothesis might see be true and your decision might still be in error.
4.
a.
1.96  
1.96   20
1.96   20 
 1.96
, x+
,50 +
 x   50    50  7.84,50  7.84

n
n  
25
25 
  42.16,57.84
b.

s
s  
20
20 
, x  t1  ,n 1
,50  2.064
   50  8.256,50  8.256
 x  t1  ,n 1
   50  2.064
2
2

n
n 
25
25 
  41.74,58.26
5.
a.
Let = mean price of Civics in San Francisco. The two hypotheses are:
H0 :  = 8500
Ha :   8500
We are assuming that the distribution of Honda Civics is normal.
b.
The alternative is two-sided, because the hypothesis is that  differs from the national average, in
other words either greater than or less than.
c.
The acceptance region for the null hypothesis covers the range

 

z
1  / 2

n
, 
 

n
z
1  / 2
where =600, n=9, =8500, and =0.05. Finally, use Excel’s NORMSINV function to calculate
the value of z1–/2=NORMSINV(0.975)=1.96. The acceptance region is therefore

600
,8500 
 8500  1.96

9
z
1  / 2
600 
   8500  392,8500  392   8108,8892
9
Since the acceptance region does not include the sample mean of 9000, we reject the null
hypothesis and accept the alternative that the mean price of Civics in San Francisco is not equal to
$8500.
1
Chapter 6: Statistical Inference
d.
One way to perform this test is with a one-sample t test. The t statistic is :
t
x 
s n
where x  9000 , =8500, s=600, and n=10. So that t  9000  8500  2.635
600 10
Using Excel’s TDIST function, TDIST(2.635,9,2)=0.0271. Hence, the probability of the t statistic
is 2.71%, and at the 5% significance level, the price difference is significant.
6.
a.
The unknown population mean of the American speakers is μ1, the population mean for the
imported speakers is μ2. The null and alternate hypothesis are:
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
b.
Call the sample average of the American speakers x1, and the sample average of the imported
speakers x2. The unknown population mean of the American speakers is 1, the population mean for
the imported speakers is 2To test whether these averages indicate that the types of speakers are
different in price, use the t-test:



t   x x 1 1   where s  n 1 s  n
1
2
s
1
2
 1 s 22
n1  n2  2

n n
1
2
1
1
2
2
The first group, the American-made speakers, has a sample size, n1 = 10 and a standard deviation,
s1 =5. The second group, the imported speakers has a sample size, n2 , equal to 5 with a standard
deviation, s2 = 4. Therefore, the pooled standard deviation, s, is:
10 1  5 1
s   5   4
2
10  5  2
2
=
10 152  5 1 42 
10  5  2
289
 4.715
13
Under the null hypothesis, the value of the t statistic is:



90  85  0
t   x x 1 1   = 
=
1
2
s
1
2

n n
1
2
4.715 1  1
10 5
5  1936
.
2.5825
This t statistic follows the t distribution with 13 degrees of freedom under the null hypothesis.
Using Excel’s TDIST function we can calculate the probability of a t distributed random variable
having such an extreme value. The form of the Excel function is TDIST(1.936,13,2) = 0.0749.
The difference between the two speakers is not significant at the 5% level and we cannot reject the
hypothesis of equal population means for the two speaker groups.
c.
It is a significant difference at the 10% level, so at this lower level of statistical significance, we can
reject the hypothesis of equal means.
2
Chapter 6: Statistical Inference
7.
Given the distribution of the t-statistic, we know that:
x


P  t1 / 2,n1 
 t1 / 2,n1   1  
/ n


Multiplying the inequalities by
/ n
yields:

 

P  t1 / 2,n1
 x    t1 / 2,n1
  1
n
n

and subtracing x from each term of this inequality, we get:

 

P  x  t1 / 2,n1
     x  t1 / 2,n1
  1
n
n

We multiply each term by –1 (changing the signs and the direction of the inequalities):

 

P x  t1 / 2,n1
   x  t1 / 2,n1
  1
n
n

and finally, we rearrange the terms and inequalities to arrive at the final form of the confidence interval:

 

P x  t1 / 2,n1
   x  t1 / 2,n1
  1
n
n

So, the upper and lower confidence limits for μ are:
x  t1 / 2,n1

n
8.
a.
Let μ1 = mean number of beds for non-rural homes and μ2 = mean number of beds in the rural
homes. The null and alternative hypothese are:
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
3
Chapter 6: Statistical Inference
The results of the two-sample t-test are:
b.
Descriptive Statistics
N
Beds
Mean
Std. Dev
Std. Err
Location = "Non-rural"
18
111.39
43.568
10.269
Location = "Rural"
34
83.68
36.436
6.249
t test Analysis (pooled)
Mean Diff.
Beds
Std. Err.
27.71
11.370
t
df
2.437
p-value
50.00
0.018
lower
95%
4.87
upper
95%
50.55
t test Analysis (unpooled)
Mean Diff.
Beds
Std. Err.
27.71
Equality of Variance Tests
c.
12.021
t
df
p-value
2.305
29.81
0.028
F-test
Barlett
Levene
0.370
0.395
0.808
lower
95%
3.13
upper
95%
52.30
The distribution for the two location types appears as follows:
The standard deviation of the two samples is pretty close (43.568 and 36.436), which would lead us
to believe that a pooled estimate of the standard deviation is the way to go. Moreover, the results of
the F-test, Bartlett test, and Levene test, support this conclusion. However there is an outlier in both
samples, which may cause us to doubt the validity of using the t-test in this situation.
4
Chapter 6: Statistical Inference
In the Mann-Whitney test, we choose between the following hypotheses:
d.
H0: The median number of beds in the rural nursing homes = median number of beds in the nonrural homes.
Ha: The median number of beds are not equal.
The results of the Mann-Whitney test are:
N
Beds
3rd Quartile
Maximum
"Non-rural"
18
Minimum
60.00
1st Quartile
81.75
Median
120.00
120.00
244.00
"Rural"
34
25.00
59.25
80.00
106.50
221.00
Mann-Whitney Rank Analysis
Beds
Median Diff.
Rank Sum1
26.50
613.5
Rank Sum2
p-value
764.5
lower 95%
0.009
upper 95%
4.00
52.00
Based on the results of the text, we reject the null hypothesis with a p-value of 0.009 and accept the
alternative hypothesis that there are more beds in non-rural homes.
e.
Whether we use the results of the t-tests or the Mann-Whitney test, the conclusion is the same: there
are significantly more beds in non-rural homes. The confidence intervals and estimates of the
difference between the locations are also very similar.
a.
The first few values of the Days_Beds variable are:
9.
Beds
Revenue
Salaries
Expenses
244
128
385
23521
5230
5334
59
155
203
9160
2459
493
120
281
392
21900
6304
120
291
419
22354
6590
120
238
363
17421
65
180
234
120
306
90
214
96
120
b.
Medical Days
Total Days
Location
Days Beds
Non-rural
1.578
Rural
3.441
6115
Non-rural
3.267
6346
Non-rural
3.492
5362
6225
Non-rural
3.025
10531
3622
449
Rural
3.600
372
22147
4406
4998
Rural
3.100
305
14025
4173
966
Rural
3.389
155
169
8812
1955
1260
Non-rural
1.760
133
188
11729
3224
6442
Rural
1.567
First use a two sample t-test. Let μ1 = mean of the Days_Beds variable for non-rural homes and μ2 =
mean of the Days_Beds variable for rural homes. The null and alternative hypotheses are:
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
Another approach is to use the Mann-Whitney test to evaluate the hypotheses:
H0: The median Days_Beds value in the rural nursing homes = median Days_Beds value in the nonrural homes.
Ha: The median Days_Beds values are not equal.
5
Chapter 6: Statistical Inference
The results of the two tests are:
Descriptive Statistics
N
Days Beds
Std. Dev
Std. Err
Location = "Non-rural"
18
3.01505
Mean
0.576282
0.135831
Location = "Rural"
34
3.11144
0.637401
0.109313
t test Analysis
Mean Diff.
Days Beds
-0.09640
Std. Err.
0.179938
t
df
-0.536
50.00
p-value
0.595
lower 95%
upper 95%
-0.45781
0.26502
Equality of Variance Tests
F-Test
Bartlett
0.673
Descriptive Statistics
1st
Minimum
Quartile
Median
N
Days Beds
Levene
0.641
3rd
Quartile
0.688
Maximum
Location = "Non-rural"
18
1.578
2.775
3.222
3.458
3.550
Location = "Rural"
34
1.567
2.818
3.279
3.545
4.699
Median Diff.
Days Beds
-0.061
Mann-Whitney Rank Analysis
Rank
Rank
Sum1
Sum2
p-value
lower
95%
435.500
-0.322
942.500
0.419
upper
95%
0.165
c.
The distribution of the Days_Beds variable appears as follows:
d.
We fail to reject the null hypothesis under either test. There is no indication that the beds are being
utilized at different rates.
a.
Let μ1 = mean draft number in the first half of the year and μ 2 = mean draft number in the second
half of the year. The null and alternative hypotheses are:
10.
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
6
Chapter 6: Statistical Inference
b.
There is not a significant statistic for rejecting the hypothesis that the standard deviation of the two
samples are equal, so we'll use the pooled estimate. The result of the two-sample t-test is:
Descriptive Statistics
N
Number
Mean
Std. Dev
Std. Err
Half = 1
182
206.38
106.149
7.868
Half = 2
184
160.92
100.757
7.428
t test Analysis
Mean Diff.
Number
45.46
Std. Err.
10.817
t
df
4.202
364.00
p-value
0.000
lower 95%
upper 95%
24.18
66.73
Equality of Variance Tests
F-Test
0.482
c.
Bartlett
0.483
Levene
0.380
The distribution of the two samples appears as:
The distribution resembles a Uniform distribution, however since the t-test is robust to problems with nonNormality, it can still be used here.
d.
The 95% confidence intervals are:
N
Mean
Std. Dev
Std. Err
lower 95%
upper 95%
Half = 1
182
206.38
106.149
7.868
190.85
221.9
Half = 2
184
160.92
100.757
7.428
146.27
175.58
e.
A draft number selected for a person born in the first half of the year is, on average, 45.46 points
higher than a draft number for a person whose birthday falls in the second half of the year. This is a
statistically significant difference. The ramification of this result is that the assignment of draft
numbers is not truly random and that people born in the second half of the year are more likely to
receive low draft numbers, and hence are more likely to be drafted.
7
Chapter 6: Statistical Inference
11.
a.
Let μ1 = mean female salary and μ2 = mean male salary. The null and alternative hypotheses are:
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
We'll use a significance level of 5% for this test.
b.
The unpooled and pooled t-tests results are:
Descriptive Statistics
N
Salary
Mean
Std. Dev
Std. Err
Sex = "F"
37
27,027.35
5,478.231
900.616
Sex = "M"
44
33,004.91
5,831.991
879.206
unpooled t test Analysis
Mean Diff.
Salary
-5,977.56
Std. Err.
1,258.615
t
df
-4.749
p-value
78.00
0.000
lower 95%
upper 95%
-8,483.27
-3,471.85
pooled t test Analysis
Mean Diff.
Salary
-5,977.56
Std. Err.
1,265.517
t
df
-4.723
79.00
p-value
0.000
lower 95%
upper 95%
-8,496.51
-3,458.61
Equality of Variance Tests
F-Test
0.705
Bartlett
0.698
Levene
0.000
The conclusions of the two tests are the same, suggesting that there is a statistically significant difference
in salary between the male and female professors. A histogram of the difference appears as follows:
8
Chapter 6: Statistical Inference
c.
There are only enough data for the assistant professor and instructor groups. The unpooled and
pooled t-test results are:
Descriptive Statistics
N
asst prof
instructor
Mean
Std. Dev
Std. Err
Sex = "F"
17
28,274.18
6,598.744
1,600.430
Sex = "M"
15
31,202.33
5,027.220
1,298.023
Sex = "F"
20
25,967.54
4,197.807
938.658
Sex = "M"
17
30,960.41
5,829.235
1,413.797
unpooled t test Analysis
Mean Diff.
Std. Err.
t
df
p-value
lower
95%
upper
95%
asst prof
-2,928.16
2,060.641
-1.421
29.42
0.166
-7,142.64
1,286.33
instructor
-4,992.87
1,697.027
-2.942
28.54
0.006
-8,469.08
-1,516.66
lower
95%
upper
95%
pooled t test Analysis
Mean Diff.
Std. Err.
t
df
p-value
asst prof
-2,928.16
2,096.262
-1.397
30.00
0.173
-7,209.29
1,352.98
instructor
-4,992.87
1,652.707
-3.021
35.00
0.005
-8,348.05
-1,637.69
Equality of Variance Tests
F-Test
Bartlett
Levene
asst prof
0.312
0.307
0.173
instructor
0.173
0.177
0.005
The conclusion from the pooled and unpooled t-tests is the same: there is a significant difference for
instructors, but not one for assistant professors.
d.
There is some evidence that the university has underpaid its female faculty members. The overall
significance level for all ranks is less than 0.0001. When testing for individual ranks, there is
statistical significance only for the instructors. There is no significant difference for assistant
professors and not enough evidence for other teaching ranks.
However, there are some factors that have not been considered yet. For example, the age at which a
person has been hired and a measure of the person's teaching experience should also be considered
in any analysis of this type.
12.
a.
Let μ1 = mean athlete graduation rate and μ2 = mean female graduation rate. Let μd = μ1 – μ2. The
null and alternative hypotheses are:
H0: μd = 0
H a: μ d ≠ 0
9
Chapter 6: Statistical Inference
b.
The results of the paired t-test are:
Descriptive Statistics
N
Mean
Std. Dev.
Std. Err.
White Females
11
82.73
7.254
2.187
White Males
11
68.45
8.466
2.553
t-Test Analysis
Difference
N
Mean
Std. Dev
11
14.27
6.513
Std. Err
t
1.964
df
7.268
lower
95%
p-value
10
0.000
upper
95%
9.90
18.65
There is strong evidence for a difference in the graduation rates. White female athletes have a
graduation rate that is 14.27 points higher than their male counterparts. The 95% confidence
interval for the difference is (9.90, 18.65). The 90% confidence interval is (10.71, 17.83).
c.
The results of the 1 sample Wilcoxon test are:
Descriptive Statistics
N
Minimum
1st Quartile
Median
3rd Quartile
Maximum
White Females
11
71
78
82
87
94
White Males
11
55
64
69
72
83
Wilcoxon Sign Rank Analysis
N
Difference
N<0
11
N=0
0
N>0
0
Median
11
p-value
15.00
0.001
lower
95%
upper
95%
10.00
19.00
The results of the Sign test are:
N
White
Females
White
Males
Minimum
Descriptive Statistics
1st
3rd
Quartile
Median
Quartile
Maximum
11
71
78
82
87
94
11
55
64
69
72
83
Sign Test Analysis
N
Difference
11
N<0
N=0
0
N>0
0
Median
11
15.00
p-value
0.001
lower
95%
9.28
upper
95%
20.15
Achieved
Confidence
0.950
In both tests we reject the null hypothesis, accepting the alternative that there is a difference in
graduation rates.
d.
We make the assumption that the only difference involved is gender since each pair comes from the
same school and involves athletes, so these are paired comparsions.
e.
White female athletes have a significantly higher graduation rates than white male athletes in the
Big Ten conference. However since this sample was not a random sample but a sample of a specific
group, the results can only be applied to the Big Ten schools. To test whether these results can be
applied to all universities, a random sample of the universities will have to be created.
a.
Let μ1 = mean white refusal rate and μ2 = mean minority refusal rate. Let μd = μ1 – μ2. The null and
alternative hypotheses are:
13.
H0: μd = 0
H a: μ d ≠ 0
10
Chapter 6: Statistical Inference
The paired t-test results are:
b.
Descriptive Statistics
N
Mean
Std. Dev.
Std. Err.
Minority
20
36.882
13.0509
2.9183
White
20
15.625
7.7958
1.7432
t-Test Analysis
Difference
N
Mean
Std. Dev
Std. Err
t
20
21.257
8.2954
1.8549
df
11.460
p-value
19
0.000
lower
95%
upper
95%
17.374
25.139
The histogram and P-plot appears as follows:
c.
6.0
5.0
4.0
3.0
2.0
1.0
0.0
6.2
8.7
11.1
13.5
16.0
18.4
20.9
23.3
25.7
28.2
30.6
33.1
35.5
37.9
40.4
Count
Differences in Refusal Rates
Refusal Rates
P-Plot of Difference in Refusals
1.132
0.132
-0.868
-1.868
5.0
15.0
25.0
35.0
There is no evidence of a violation of the assumption of Normality.
d.
The results of the Wilcoxon Signed Rank test are:
Descriptive Statistics
Minimum
1st Quartile
Median
3rd Quartile
Maximum
Minority
N
20
10.6
26.4
37.4
45.3
62.2
White
20
3.7
9.2
15.8
20.2
32.4
Wilcoxon Sign Rank Analysis
N
Difference
20
N<0
N=0
0
N>0
0
20
11
Median
19.050
p-value
0.000
lower 95%
upper 95%
17.315
25.100
Chapter 6: Statistical Inference
The results of the Wilcoxon test match the results of the t-test.
For high income applicants, the results of the t-test and the Wilcoxon test are as follows:
e.
Descriptive Statistics
N
Mean
Std. Dev.
Std. Err.
High Income Minority
20
27.515
10.9220
2.4422
High Income White
20
11.300
6.5164
1.4571
t-Test Analysis
Difference
N
Mean
Std. Dev
Std. Err
20
16.215
8.1082
1.8131
N
Minimum
t
pvalue
df
8.943
19
Descriptive Statistics
1st
3rd
Quartile
Median
Quartile
0.000
lower
95%
upper
95%
12.420
20.010
Maximum
High Income Minority
20
5.8
21.3
29.1
37.0
41.3
High Income White
20
2.2
7.4
9.8
15.1
26.8
Wilcoxon Sign Rank Analysis
N
Difference
N<0
20
N=0
0
N>0
1
19
Median
17.350
p-value
0.000
The distribution of the paired differences in refusal rates appears as follows:
Differences in Refusal Rates
(High Income)
6.0
Counts
5.0
4.0
3.0
2.0
1.0
3.0
5.1
7.1
9.1
11.1
13.2
15.2
17.2
19.3
21.3
23.3
25.3
27.4
29.4
1.0
0.0
Refusal Rates
P-Plot of Difference in Refusals
(High Income)
1.132
0.132
-0.868
-1.868
0.0
10.0
20.0
30.0
There is no reason to doubt the assumption of normality from these charts.
12
lower
95%
upper
95%
12.100
20.750
Chapter 6: Statistical Inference
14.
a.
The 95% confidence intervals for the numeric variables are:
N
Salary Pupil Ratio
Spending per Pupil
Teacher Salary
b.
Std. Dev.
lower 95%
upper 95%
Area = "North"
21
Mean
6.3510
0.78299
5.9945
6.7074
Area = "South"
17
7.1318
0.83919
6.7003
7.5632
Area = "West"
13
7.1131
1.50153
6.2057
8.0204
Overall
51
6.8055
1.07669
6.5027
7.1083
Area = "North"
21
3,900.62
796.898
3,537.88
4,263.36
Area = "South"
17
3,274.41
756.910
2,885.24
3,663.58
Area = "West"
13
3,919.15
1,560.191
2,976.34
4,861.97
Overall
51
3,696.61
1,054.761
3,399.95
3,993.26
Area = "North"
21
24,424.14
3,725.544
22,728.30
26,119.99
Area = "South"
17
22,894.00
3,553.857
21,066.78
24,721.22
Area = "West"
13
26,158.62
5,123.734
23,062.37
29,254.86
Overall
51
24,356.22
4,179.426
23,180.73
25,531.70
The 95% non-parametric confidence intervals are:
N
Salary Pupil Ratio
Spending per Pupil
Teacher Salary
Median
lower 95%
upper 95%
Area = "North"
21
6.25
5.960
6.710
Area = "South"
17
7.34
6.660
7.585
Area = "West"
13
6.86
6.055
8.070
Overall
51
6.77
Area = "North"
21
3,621
3,457.0
4,278.0
Area = "South"
17
2,980
2,828.5
3,664.5
Area = "West"
13
3,705
3,115.5
4,603.0
Overall
51
3,554
3,334.0
3,840.0
Area = "North"
21
24,500
22,632.5
26,406.0
Area = "South"
17
22,080
21,179.5
24,003.0
Area = "West"
13
25,788
23,561.0
27,488.5
Overall
51
23,382
23,029.5
25,036.0
6.455
7.090
15.
a.
The null and alternative hypotheses are:
H0: The number of pollution days is the same throughout the time period.
Ha: The number of pollution days is not equal.
b.
The results of the t-test analysis are:
t-Test Analysis
N
Diff80
14
Mean
-16.600
Std. Dev.
26.2461
Std. Err.
7.0146
t
2.367
df
13.000
p-value
0.034
lower 95%
-31.754
upper 95%
-1.446
Based on this analysis we reject the null hypothesis and conclude, with a p-value of 0.034, that the
average number of pollution days from 1985 to 1989 is significantly lower than the number of
days in 1980. The number of pollution days decreased from 1.446 to 31.754 days.
13
Chapter 6: Statistical Inference
The Normal P-plot appears as follows:
c.
1.292
0.792
0.292
-0.208
-0.708
-1.208
-1.708
-99.6
-79.6
-59.6
-39.6
-19.6
0.4
Because of one extreme outlier there is some question whether the t-test is appropriate for this
data. The outlier might have a large impact on the test's conclusion.
d.
The results of the Wilcoxon test are:
Wilcoxon Sign Rank Analysis
N
Diff80
14
N<0
N=0
N>0
11
0
3
Median
p-value
-10.9
0.008
lower 95%
-21.600
upper 95%
-4.400
The test results are actually stronger than those provided by the t-test. This is probably due to the
fact that the Wilcoxon test is not influenced by the presence of a large outlier. Because of the
outlier, it is better to report the non-parametric result.
16.
a.
The null and alternative hypotheses are:
H0: The resistance levels will the same for men and women throughout the course of the study.
Ha: The resistance levels will not be the same.
14
Chapter 6: Statistical Inference
b.
The descriptive statistics of the Resistance variable are:
N
Resistance
Day= 1
Resistance
Day= 5
Resistance
Day= 9
Resistance
Day= 10
Resistance
Day= 13
Resistance
Day= 16
Resistance
Day= 19
Resistance
Day= 20
Resistance
Resistance
Day= 22
Day= 24
Mean
Std. Dev
Std. Err
Gender = "Female"
8
211.400
10.0504
3.5534
Gender = "Male"
7
185.071
21.0666
7.9624
Gender = "Female"
8
211.025
16.5697
5.8583
Gender = "Male"
7
179.871
20.5025
7.7492
Gender = "Female"
8
203.763
16.0285
5.6669
Gender = "Male"
7
175.314
17.3432
6.5551
Gender = "Female"
8
254.463
15.5860
5.5105
Gender = "Male"
7
199.829
24.1288
9.1198
Gender = "Female"
8
238.600
14.6502
5.1796
Gender = "Male"
7
195.629
19.6891
7.4418
Gender = "Female"
8
241.975
21.7236
7.6805
Gender = "Male"
7
202.100
21.0396
7.9522
Gender = "Female"
8
236.750
15.4489
5.4620
Gender = "Male"
7
200.414
18.1770
6.8703
Gender = "Female"
8
208.163
23.2408
8.2169
Gender = "Male"
7
179.071
14.6212
5.5263
Gender = "Female"
8
191.838
22.5575
7.9753
Gender = "Male"
7
182.386
19.4655
7.3573
Gender = "Female"
8
192.750
19.2699
6.8129
Gender = "Male"
7
179.414
13.1021
4.9521
The t-test results are:
Mean Diff.
Resistance
c.
Std. Err.
t
df
p-value
lower
95%
upper
95%
Day= 1
26.329
8.3327
3.160
13.00
0.008
8.327
44.330
Day= 5
31.154
9.5690
3.256
13.00
0.006
10.481
51.826
Day= 9
28.448
8.6163
3.302
13.00
0.006
9.834
47.062
Day= 10
54.634
10.3447
5.281
13.00
0.000
32.286
76.982
Day= 13
42.971
8.8815
4.838
13.00
0.000
23.784
62.159
Day= 16
39.875
11.0811
3.598
13.00
0.003
15.936
63.814
Day= 19
36.336
8.6758
4.188
13.00
0.001
17.593
55.079
Day= 20
29.091
10.2144
2.848
13.00
0.014
7.024
51.158
Day= 22
9.452
10.9651
0.862
13.00
0.404
-14.237
33.140
Day= 24
13.336
8.6475
1.542
13.00
0.147
-5.346
32.018
There are significant differences between males and females on every day except days 22 and 24.
This is the case even if an unpooled variance estimate is used.
15
Chapter 6: Statistical Inference
The scatterplot appears as follows:
d.
Resistance Data
300
280
Resistance
260
240
Female
Male
220
200
180
160
140
0
5
10
15
20
25
Day
e.
The results of the Mann-Whitney test are:
Median Diff.
Resistance
Rank Sum1
Rank Sum2
p-value
lower
95%
upper
95%
Day= 1
-31.250
38.5
81.5
0.047
-44.400
0.000
Day= 5
-32.700
35.0
85.0
0.014
-52.000
-9.700
Day= 9
-29.900
33.0
87.0
0.006
-48.500
-7.400
Day= 10
-63.200
29.0
91.0
0.001
-75.800
-30.600
Day= 13
-43.050
30.0
90.0
0.001
-56.900
-22.200
Day= 16
-44.750
33.0
87.0
0.006
-62.600
-14.500
Day= 19
-39.950
32.0
88.0
0.004
-56.500
-16.200
Day= 20
-27.650
36.5
83.5
0.025
-55.800
-5.800
Day= 22
-9.850
48.5
71.5
0.430
-36.200
15.900
Day= 24
-15.100
42.0
78.0
0.121
-32.000
6.600
The conclusions of this test match the conclusions of the t-test.
f.
During the first nine days of observation–the control period–women had a higher resistance scores
than males, indicating a higher level of blood loss. During the test period–from day 10 through day
20–the resistance levels increased. This was most evident in the first day of the test period, and was
particularly noticeable for females. In the recovery period–day 22 and 24–the resistance levels
decreased and there was no significant difference between the male and female participants.
a.
Let μ1 = mean exam score for control students and μ2 = mean exam score for experimental method
students. This is a two-sided test. The null and alternative hypotheses are:
17.
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
16
Chapter 6: Statistical Inference
b.
The results of the unpooled and pooled t-tests are:
Descriptive Statistics
N
Final Score
Std. Dev
Std. Err
Group = "Control"
79
Mean
23.87
11.597
1.305
Group = "Experimental"
85
27.34
8.847
0.960
unpooled t test Analysis
Mean Diff.
Final Score
Std. Err.
-3.47
1.620
t
-2.141
df
145.64
p-value
0.034
lower
95%
-6.67
upper
95%
-0.27
pooled t test Analysis
Mean Diff.
Final Score
Std. Err.
-3.47
1.604
t
-2.162
df
162.00
p-value
0.032
lower
95%
-6.64
upper
95%
-0.30
Equality of Variance Tests
F-Test
0.015
Bartlett
0.016
Levene
0.039
Based on the results of the equality of variance tests, we should use the unpooled t-test results.
Based on that test, the 95% confidence interval for the final score is (–6.67, –0.27) with a p-value
of 0.034. Test scores are statistically significantly higher for students using the experimental
method.
c.
The distribution of scores for the two groups are:
The scores are not evenly distributed. The distribution of the scores in both methods seems to
break into two groups. The first group contains those students who score less than 10 on the final
exam. The second group consists of those students who score 19 or better on the final. There are
very few students who score between 10 and 19. Because of the shape of the distribution, there is
some question whether the t-test is the appropriate test for this data.
17
Chapter 6: Statistical Inference
d.
The results of the Mann-Whitney test are:
Descriptive Statistics
N
Final Score
Minimum
1st Quartile
Median
3rd
Quartile
Maximum
Group = "Control"
79
0.00
20.00
28.00
32.00
39.00
Group = "Experimental"
85
0.00
25.00
29.00
34.00
38.00
Mann-Whitney Rank Analysis
Median Diff.
Final Score
-2.00
Rank Sum1
Rank Sum2
6,006.5
7,523.5
lower
95%
p-value
0.092
upper
95%
-4.00
0.00
Using the non-parametric test, we would not reject the null hypothesis since the p-value is 0.092.
e.
We arrive at two different conclusions based on the two statistical tests. In examining the
distribution of the scores, we noticed that scores did not appear to follow the Normal distribution.
This causes us to have some concern about the appropriateness of the t-test and fail to reject the null
hypothesis based on the results of the Mann-Whitney test.
a.
The results of the paired t-tests are:
18.
Descriptive Statistics
Region = "E"
Region = "MW"
Region = "S"
Region = "W"
Overall
N
Mean
Std. Dev.
Std. Err.
Dem1980
13
42.415
5.4626
1.5150
Dem1984
13
42.162
5.1699
1.4339
Dem1980
10
37.690
7.1507
2.2612
Dem1984
10
40.040
6.5862
2.0827
Dem1980
10
48.000
4.1918
1.3256
Dem1984
10
38.370
2.0022
0.6332
Dem1980
17
34.106
6.8112
1.6520
Dem1984
17
35.753
6.0693
1.4720
Dem1980
50
39.762
7.9227
1.1204
Dem1984
50
38.800
5.8179
0.8228
t-Test Analysis
N
Mean
Std. Dev
Std. Err
t
df
pvalue
lower
95%
upper
95%
Region = "E"
Difference
13
0.254
3.5225
0.9770
0.260
12
0.799
-1.875
Region = "MW"
Difference
10
-2.350
3.3909
1.0723
-2.192
9
0.056
-4.776
0.076
Region = "S"
Difference
10
9.630
3.3549
1.0609
9.077
9
0.000
7.230
12.030
Region = "W"
Difference
17
-1.647
3.6797
0.8925
-1.846
16
0.084
-3.539
0.245
Overall
Difference
50
0.962
5.6308
0.7963
1.208
49
0.233
-0.638
2.562
b.
2.382
The only area in which there was a significant difference in voting was the Southern region. The pvalue for this region is < 0.0001 and the 95% confidence interval is (7.23, 12.03) indicating a
decline in voting percentage from 7 to 12 points. Overall, there was not a significant difference
after pooling all regions.
18
Chapter 6: Statistical Inference
19.
a.
Let μ1 = mean exam score for the female students and μ2 = mean exam score for male students. The
null and alternative hypotheses are:
H0: μ1 – μ2 = 0
H a: μ 1 – μ 2 ≠ 0
b.
The result of the two-sample t-tests are:
Descriptive Statistics
N
Exam Score
Mean
Std. Dev
Std. Err
Gender = "Female"
37
80.41
10.057
1.653
Gender = "Male"
43
79.93
12.618
1.924
unpooled t test Analysis
Mean Diff.
Exam Score
Std. Err.
0.48
2.537
t
df
0.187
77.58
p-value
0.852
lower
95%
-4.58
upper
95%
5.53
pooled t test Analysis
Mean Diff.
Exam Score
Std. Err.
0.48
2.580
t
df
0.184
78.00
p-value
0.854
lower
95%
-4.66
upper
95%
5.61
Equality of Variance Tests
F-Test
0.167
Bartlett
0.165
Levene
0.193
There is no reason not to use the pooled t-test results. We fail to reject the null hypothesis with a
p-value of 0.854. The 95% confidence interval for the difference in scores is (–4.66, 5.61).
c.
The distribution of the scores for the two groups is:
The t-distribtuion assumes a continuous distribution from minus infinity to positive infinity,
however the exame scores are constrained to the range [0, 100]. In practical terms, the range is
more like (70, 100]. Moreover exam scores are discrete, only taking on integer values. However,
the robustness of the t-test to departures from normality may allow us to still the analyze the data
with the t-test without coming to an erroneous conclusion.
19
Chapter 6: Statistical Inference
20.
The t-test results are:
a.
React1 vs. React2 t-Test Analysis
Difference
N
Mean
Std. Dev
Std. Err
14
0.00350
0.020843
0.005570
N
Mean
Std. Dev
Std. Err
14
0.01429
0.016198
0.004329
N
Mean
Std. Dev
Std. Err
14
0.01079
0.019589
0.005235
t
df
0.628
p-value
13
lower 95%
upper 95%
-0.00853
0.01553
lower 95%
upper 95%
0.00493
0.02364
lower 95%
upper 95%
-0.00052
0.02210
0.541
React1 vs. React3 t-Test Analysis
Difference
t
df
3.300
p-value
13
0.006
React2 vs. React3 t-Test Analysis
Difference
t
2.060
df
p-value
13
0.060
As the meet continues, the reaction time drops with each round. The drop is not significant between round
1 around round 2, but it is significant when the first round is compared with the third round and almost
significant when the second round is compared with the third.
The P-plots for each paired difference are:
b.
React1 vs. React3
React1 vs. React2
1.292
1.292
0.792
0.792
0.292
0.292
-0.208
-0.208
-0.708
-0.708
-1.208
-1.208
-1.708
-0.035
-0.015
0.005
-1.708
-0.008
0.025
0.002
0.012
0.022
0.032
0.042
React2 vs. React3
1.292
0.792
0.292
-0.208
-0.708
-1.208
-1.708
-0.016
0.004
0.024
0.044
There is no reason to reject an assumption for normality for the 1 to 2 plot and for the 1 to 3 plot;
however there are some problems with the 2 to 3 plot. One value appears out of line with the
others, making it appear as if the data were not normally distributed.
20
Chapter 6: Statistical Inference
c.
The Wilcoxon test results are:
React1 vs. React2 Wilcoxon Sign Rank Analysis
N
Difference
14
N<0
N=0
5
N>0
1
8
Median
p-value
0.00750
0.542
lower 95%
upper 95%
-0.00800
0.01700
React1 vs. React3 Wilcoxon Sign Rank Analysis
N
Difference
14
N
Difference
14
N<0
N=0
1
N<0
N>0
2
N=0
4
11
N>0
0
10
Median
p-value
0.01100
lower 95%
0.003
Median
p-value
0.01000
0.078
upper 95%
0.00400
0.02450
lower 95%
upper 95%
-0.00050
0.02100
The results match what was observed using the t-test: 1) There is no significant difference
between round 1 and round 2, 2) There is a significant difference between round 1 and round 3,
and 3) There is no significant difference between round 2 and round 3 at the 5% level, but there is
one at the 10% level.
d.
The reaction times decrease as the sprinters advance in the competion. There is not a significant
decrease in the reaction times between the first and second round, but comparing the third round
reaction times to those in the first show a significant decrease of 0.014 seconds.
a.
The t-test results are:
21.
Round1 vs. Round2 t-Test Analysis
Difference
N
Mean
14
0.1043
Std. Dev
0.12233
Std. Err
0.03269
t
df
3.190
p-value
13
0.007
lower
95%
upper
95%
0.0337
0.1749
lower
95%
upper
95%
-0.0284
0.0298
Round2 vs. Round3 t-Test Analysis
Difference
b.
N
Mean
14
0.0007
Std. Dev
0.05045
Std. Err
0.01348
t
0.053
df
p-value
13
0.959
There is a significant decrease in the race time between the first and second round with a p-value of
.007. The 95% confidence interval for the difference is (0.0337, 0.1749). However there is no
significant difference between the times in round 2 and round 3. This suggests that the level of
competition in the second and third rounds is high enough that the race times do not decrease as
much. The competition in the first round is not as severe, allowing the elite racers to conserve their
strength for the later rounds.
21
23