Name: ANSWER KEY
1.
(True of
) A nonparametric test is one that assumes that the population from which a sample is drawn is normally distributed or approximately so.
2.
(True of
) Compared to a parametric test, nonparametric tests use the information in the data more efficiently.
3.
(
of False) Analysis of variance (ANOVA) is a set of techniques that allow us to compare three or more sample means at the same time.
4.
(
of False) Comparing the variation between the groups to the variation within the groups is the basis of ANOVA.
5.
(
of False) An assumption of the ANOVA model is that the population distributions are normal.
6.
The F-statistic in a one-way ANOVA represents the: a. variation between the treatments plus the variation within the treatments. b. variation within the treatments minus the variation between the treatments. c.
. d. variation within the treatments divided by the variation between the treatments.
7.
A study from Harvard School of Public Health shows that men who have gum disease are at high risk for having blocked arteries in their legs. This study shows that men who suffer from gum disease
have one and a half times the chance of having blocked arteries to their legs, while losing teeth from gum disease doubles chances. Having lost a tooth in the last six years markedly increased a person's chances of having blocked arteries.
Is this an example of a(n): a.
Designed experiment
8.
A political analyst
in Texas surveys a random sample of registered Democrats and compares the results with those obtained from a random sample of registered Republicans. This would be an example of: a.
. b. Dependent samples. c. Independent samples only if the sample sizes are equal. d. Dependent samples only if the sample sizes are equal.

Name: ANSWER KEY
9.
To test the effectiveness of three different types of lawn fertilizer, each type was applied to four different lawns. The same size patch of lawn was mowed for all 18 test units and the grass clippings were weighed. The weight in pounds are shown in the following table.
Lawn 1
Fertilizer 1
10.3
Fertilizer 2
11.7
Fertilizer 3
8.2
Lawn 2
Lawn 3
Lawn 4
Mean
8.6
8.2
10.6
9.4
12.0
9.3
10.5
10.9
8.9
10.7
9.3
9.3
Using the above table, identify the following: a.
The independent variables:
b.
The dependent variables:
c.
The value of x
2,3
:
d.
The value of x̅
1
:
10.
My fantasy football league is divided into 2 divisions. My wife’s team is in the Shirts Division and my team is in the Skins Division. Because we play more games against opponents in our own division, my wife frequently complains that she has it tougher because the Shirts Division has the better teams. Here is a random sample of scores from each division for the last three seasons. (This information is in the Fantasy_Football data set).
Skins
1154
1502
1296
Shirts
1488
1346
1400
1116
1236
1224
1023
1029
912
841
1282
1288
1122
1137
1161
1090
1068

Name: ANSWER KEY
(Question 10, continued) Assuming equal variances, perform an appropriate t-test to determine if mean score in the Shirts division is higher than the mean score in the Skins division. Use a 90% confidence level (α=.10). t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference df t Stat
P(T<=t) one-tail t Critical one-tail
P(T<=t) two-tail
Skins
12
29574.41667
Shirts
1134.25 1237.416667
31258.38636 27890.44697
12
0
22
-
1.469457163
0.077932525
1.321236742
0.15586505 t Critical two-tail 1.717144374 a.
State your null hypothesis. H
0
: The mean score of the Shirts division is equal to or greater than the mean score of the Skins division. 𝝁
𝑺𝒌𝒊𝒏𝒔
− 𝝁
𝑺𝒉𝒊𝒓𝒕𝒔
≥ 𝒐𝒓 = 𝟎 b.
State your alternative hypothesis. H a
: The mean score of the Shirts division is greater than the mean score of the Skins division. 𝝁
𝑺𝒌𝒊𝒏𝒔
− 𝝁
𝑺𝒉𝒊𝒓𝒕𝒔
< 0 c.
State your conclusion: Because we’re testing to see if one score is greater than the other, this is a one-tailed test. Our p-value of .078 is less than our alpha of .10. In addition, our test statistic of 1.47 is larger than our one-tail critical value of 1.32. Both of these give us sufficient evidence to reject the null hypothesis and conclude that the mean score in the
Shirts division is higher than the mean score in the Skins division.
11.
Driving records indicate that young women typically have a better driving record than young men.
An insurance company wants to verify this data and surveys then number of speeding tickets received by drivers between the ages of 18 and 25 over the last year. The results are summarized, below.
Sample size
Men
120
Women
85
Sample mean
Sample standard deviation
1.2
24.8
0.4
10.6

Name: ANSWER KEY
Assuming unequal variances, conduct an appropriate t-test to determine if young women have fewer speeding tickets than young men. Use a 99% confidence level for your conclusion (α=.01).
We have summarized data for this problem, so we will need to use the Excel template for a 2 sample t-test with unequal variances.
2-Sample t-Test Comparing Means, Unequal Variances (based on summary statistics)
Hypothesized Difference (mu1 - mu2) 0.000
Summary of Sample Data:
Mean for Sample 1, xbar1 1.200
Std. Dev. For Sample 1, s1 24.800
Size of Sample 1, n1 120
Mean for Sample 2, xbar2 0.400
Std. Dev. For Sample 2, s2 10.600
Size of Sample 2, n2 85
Calculated
Values: d.f. = 172.08
Std. Error
= 2.53914 t = 0.315
p-value if the test is:
Two-
Left-Tail Tail
Right-
Tail
0.6235 0.7531 0.3765
Confidence level desired 0.90
Lower confidence limit -3.40
Upper confidence limit 5.00 a.
State your null hypothesis. H
0
: The mean number of tickets issued to women is greater than or equal to the mean number of tickets issued to men. 𝝁 𝒎𝒆𝒏
− 𝝁 𝒘𝒐𝒎𝒆𝒏
≤ 𝒐𝒓 = 𝟎 b.
State your alternative hypothesis. H a
: The mean number of tickets issued to women is less than the mean number of tickets issued to men. 𝝁 𝒎𝒆𝒏
− 𝝁 𝒘𝒐𝒎𝒆𝒏
> 𝟎 c.
State your conclusion: This is also a one-tailed test and, since our alternative hypothesis is
“greater than,” we are looking at the right tail. Since our p-value of .375 is greater than our alpha of .01, we do not have sufficient evidence to reject the null hypothesis and conclude that women have fewer speeding tickets than men.
12.
Using the data from the speeding tickets question (question 11), construct a 99% confidence interval for the mean difference in the mean number of speeding tickets for men and women.
From the above Excel results:
Confidence level desired 0.99

Name: ANSWER KEY
Lower confidence limit -3.4
Upper confidence limit 5.00
Since the confidence interval does not contain the value 0, we have further evidence to reject the null hypothesis (this was not part of the answer).
13.
A new diet program claims that clients will lose weight after 7 days on the program. To validate their claim, they recorded the starting and ending weight of 11 individuals over a 7-day period. The results are shown in the following table. (This information is in the Diet data set.)
Starting Weight Ending Weight
221 218
215
206
210
204
236
214
256
240
222
239
241
212
250
245
218
230
218
229
215
220
Conduct an appropriate t-test to determine if the claim that the clients will lose weight in 7 days is accurate. Use a 95% confidence level for your conclusion (α=.05).
These are dependent samples, so we do a paired samples t-test. t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference df t Stat
P(T<=t) one-tail t Critical one-tail
P(T<=t) two-tail t Critical two-tail
Starting
Weight
11
0.952953645
Ending
Weight
226.9090909 223.9090909
213.4909091 235.0909091
11
0
10
2.140872096
0.028974853
1.812461123
0.057949707
2.228138852

Name: ANSWER KEY a.
State your null hypothesis. H
0
: The clients will not lose weight on the diet.
𝝁 𝒅
≤ 𝟎 𝒐𝒓 𝝁 𝒔𝒕𝒂𝒓𝒕
− 𝝁 𝒆𝒏𝒅
< 𝟎 b.
State your alternative hypothesis. H a
: The clients will lose weight on the diet.
𝝁 𝒅
> 0 𝑜𝑟 𝝁 𝒔𝒕𝒂𝒓𝒕
− 𝝁 𝒆𝒏𝒅
> 0 c.
State your conclusion: Another one-tailed test. Since the p-value of .0289 is less than the alpha of .05, we have sufficient evidence to reject the null hypothesis and conclude that people lose weight on the diet plan.
14.
Complete the missing values of the ANOVA table:
Source
Between groups
Within groups
SS
14.6
10.40 df
3
16
MS
4.87
0.65
Total 25 19
F
7.5
p-value
0.0024

Name: ANSWER KEY
15.
Using information from the ANOVA table you just completed (question 14): a.
How many treatments were involved in this experiment?
4 (df for between groups = p-1) b.
State your null hypothesis. H
0
: The means for all 4 treatments are equal.
c.
State your alternative hypothesis. H a
: At least one of the means is different d.
State your conclusion at the 99% confidence level (α=.01): We have sufficient evidence to reject the null hypothesis and conclude that the alternative hypothesis is true.
16.
A consumer research organization is attempting to determine whether there is any difference in mpg for fully loaded 22-foot trucks leased from three companies, A-Haul, Bertz, and Glyder. Five of these trucks are rented from each company. Each truck is driven with the same weight cargo over the same
200 mile route and the mpg recorded. The results of the test are:
This data is available in the Trucks data set.
Perform an appropriate statistical test to determine if the mpg for the 3 trucking companies is equal.
Anova: Single
Factor
SUMMARY
A-Haul
Bertz
Glyder
Groups
ANOVA
Source of Variation
Between Groups
Within Groups
Count
5
5
5
SS
28.948
34.392
Sum
20.7
28.6
37.7
Average Variance
4.14
5.72
7.54
0.783
6.052
1.763 df MS F P-value F crit
2 14.474 5.050244 0.025626 3.885294
12 2.866

Name: ANSWER KEY
Total 63.34 14 a.
State your null hypothesis. H
0
: The means for all 3 treatments are equal.
b.
State your alternative hypothesis. H a
: At least one of the means is different c.
State your conclusion at the 95% confidence level (α=.05): We have sufficient evidence to reject the null hypothesis and conclude at least one of the means is different from the others.