STAT500 HW#10_solutions
STAT500 HW#10 Solutions
1)
(15 pts; 5pts each)
a) The value of the pooled-variance t-statistic, t = -3.83.
b) The value of the non-pooled variance t’-statistic, t’ = -3.79.
c) We want to test Ho: μ1 = μ2, Ha: μ1 ≠ μ2. Both statistics (t and t’) lead one to reject the null
hypothesis (p = 0.000 and 0.000, respectively) at  = 0.01. Similarly, the p-values for
both statistics are less than  = 0.001, we would reject the null hypothesis in this case as
well. Then there is a difference in bonus percentages for gender, and the conclusion to
reject the null hypothesis is the same at both the 1% and 0.1% levels of significance,
regardless of which statistic (t or t’) is used to test the hypothesis.
2) (16 pts; 4pts each)
The data are paired since twins are “naturally” paired. So the paired t test is used.
a) The hypotheses are Ho: a - n = 0, Ha: a – n ≠ 0 (two-tailed test). From the output,
we can reject Ho (t = 4.95; P-value = 0) at  = 0.05, and conclude that the mean
final grades are different for academic versus non-academic emphasis.
b) The mean difference between final grades of the students in academic and
non-academic environments is (from the output) 3.800. The 95% confidence
interval is (2.230, 5.370). The size difference in the mean final grades is estimated
to be between 2.230 and 5.370 with 95% confidence.
c) The conditions appear to be satisfied for the use of a paired t-procedure. For
example, the normality plot suggests that the difference is normally distributed.
d) It appears that using twins in this study to control for variation in the final scores
was effective as compared to taking a random sample of 30 students in both types
of environments. Justification is provided by rejection of the null hypothesis in the
paired-procedure (controlling for variation) and a failure to reject the same
hypothesis in an independent two-sample t-procedure.
3) (16 pts; 4pts each)
a. The data are not paired. Need to determine whether to use pooled variance or separate variance.
Also both n1 and n2 are less than 30. So need to check normality assumption for both sets.
STAT500 HW#10_solutions
Normality assumptions in both cases are satisfied.
S1
 0.6  Separate variance needs to be used (by rule of thumb)
S2
The hypotheses are H0: (wide) – (narrow) = 0, Ha: (wide) – (narrow) ≠ 0 (two-tailed
test).
Using the Minitab 2-sample t procedure (do not check the box for “assume equal
variance”)
STAT500 HW#10_solutions
P-value of the test is 0.005 with “not equal” as the alternative. For tests with level
of significance >0.005, H0 is rejected. Thus, these two types of jets have different
mean noise levels.
b. Size of the difference in the mean noise level is estimated by a 98% CI.
From Minitab output, a 98% CI for the difference is (-14.79, - 1.58).
c. Since zero is not in the 98% confidence interval, we can conclude at α = 0.02
that the two types of jets do in fact have different mean noise levels.
d. The difference between the two jet noise levels is at most equal to -14.79 with
98% confidence. This represents a small percentage of the mean noise level, and
may not be of practical importance.
4) (15 pts; 5pts each)
a)
The null hypothesis is “There is no difference in the amount spent on campaigns
between males and females”, which means the difference between female campaign
expenditures and male expenditures is equal to 0. The alternative hypothesis is “Female
candidates spend less on their campaigns than male candidates”, i.e. the difference
between female campaign expenditures and male expenditures is less than 0.
Ho: μf μm  0, Ha: μf μm < 0 (one-sided left-tailed test).
We have independent samples because the samples were randomly selected, no paired
structure. Furthermore, according to the normality plot, there is no evidence that the
samples are not drawn from a normal distribution (see chart below).
STAT500 HW#10_solutions
In addition, for variance similarity, Sf / Sm = 51.95 / 61.92 = 0.84  Sf  Sm (1.414> 0.84 >
0.707)  a pooled variance can be used.
b)
We use the pooled t-test procedure in Minitab (Stat > Basic Statistics > 2-sample t… and then
clicking “Assume equal variances” box under “Options” tab, we get the following output:
_________________________________________________________________
Two-sample T for Female_1 vs Male_1
SE
N
Mean StDev Mean
Female_1 20 245.3
52.0
12
Male_1
61.9
14
20 351.0
Difference = μ (Female_1) - μ (Male_1)
Estimate for difference: -105.7
98% upper bound for difference: -67.3
T-Test of difference = 0 (vs <): T-Value = -5.85 P-Value = 0.000 DF = 38
Both use Pooled StDev = 57.1554
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STAT500 HW#10_solutions
The p-value that Minitab calculated is less than our desired significance level (0.000 <
0.02). This means that we do have sufficient evidence to conclude that women running
for office in state legislature spend less than their male counterparts.
c.
sqrt(((20-1)51.95²+(20-1)61.92²)/(20+20-2)) = 57.15
t = (245.3 - 351)/(57.15*sqrt(1/20+1/20))= -5.85
From the table at α = 0.02 and df = 38, tα/2 = 2.42857. So, 98%
1 1
+
= -105.7 ± 43.90 = (-149.60,-61.80)
20 20
The difference between the amount spent by female and male candidates is, to a
98% level of confidence, between 61.80 and 149.60.
CI = (-105.7) ± 2.42857* 57.16 *
5) (15 pts; 5pts each)
a. Null hypothesis:
Montana.
California does not have a higher mean hysterectomy cost than
Alternative hypothesis: California has a higher mean hysterectomy cost than
Montana.
The patients were selected from each group at random and are not related to each
other. There more than 30 samples in each group, so we can assume normality of the
sample means.
We can use two sample t test with unequal variances (unequal variances since
STAT500 HW#10_solutions
SMont / SCailif = 320 / 890 = 0.36).
Minitab output is :
______________________________________________________________________________________________________
Two-Sample T-Test and C
SE
Sample
N
Mean
StDev Mean
1
200
7458
320
23
2
200 12690
890
63
Difference = μ (1) - μ (2)
Estimate for difference:
-5232.0
95% upper bound for difference:
-5121.6
T-Test of difference = 0 (vs <): T-Value = -78.23
P-Value = 0.000
DF = 249
______________________________________________________________________________________________________
α = 0.05
P-Value = 0.000
0.05 > 0.000
We reject the null hypothesis at α = 0.05.
b.
95% CI for difference:
(-5363.7, -5100.3)
The 95% CI is (-5363.7, -5100.3).
c. SMont / SCailif = 320/890 = 0.360 < 0.707 (Use separate variance t-test)
6) (15 pts)
The sample size is not large, thus we check the normal assumption of the data.
STAT500 HW#10_solutions
This plot suggests that normality of both husband and wife data are ok.
We first do two-sample t-test.
Minitab two-sample t test output (use pooled variance):
________________________________________________________________
Two-Sample T-Test and CI: Husband, Wife
Two-sample T for Husband vs Wife
N Mean StDev SE Mean
Husband 10 49.7
18.9
6.0
Wife
17.4
5.5
10 47.1
Difference = μ (Husband) - μ (Wife)
Estimate for difference:
2.60
95% lower bound for difference:
-11.48
T-Test of difference = 0 (vs >): T-Value = 0.32 P-Value = 0.376 DF = 18
Both use Pooled StDev = 18.1583
__________________________________________________________________
In this result, we cannot reject H0.
STAT500 HW#10_solutions
Now we do paired-t test. Again, we start by checking the normality of the
difference of age.
This plot suggests that the differences (husband & wife) are normally distributed.
We can, therefore, proceed with a paired t-test for the mean difference (d). We
want to test Ho: d = 0, Ha: d > 0 where d = male (mean age) – female (mean age)
at α = 0.05. (Note: This is one-sided right-tailed test.) Following is the Minitab
output:
__________________________________________________________________
Paired T-Test and CI: Husband, Wife
Paired T for Husband - Wife
N Mean StDev SE Mean
Husband
10 49.70 18.92
5.98
Wife
10 47.10 17.36
5.49
Difference 10
2.60
3.57
1.13
95% lower bound for mean difference: 0.53
T-Test of mean difference = 0 (vs > 0): T-Value = 2.31 P-Value = 0.023
____________________________________________________________________________
P-value = 0.023 < 0.05, so we reject null hypothesis at α = 0.05.
This time, we conclude that the mean age of married men is higher than the mean
age of married women.
The two test results are different. The paired-t is more appropriate because the data
STAT500 HW#10_solutions
is paired.
7) (8 pts) Following is the Minitab output:
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Power and Sample Size
Paired t Test
Testing mean paired difference = 0 (versus > 0)
Calculating power for mean paired difference = difference
α = 0.02 Assumed standard deviation of paired differences = 3.57
Difference
Sample
Size
Target
Power
Actual Power
1
109
0.8
0.800122
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So, we should sample at least 109 married couples.