Elementary Statistics

Page 1 of 6
Elementary Statistics: Review for final exam Solutions
1. D.19 White blood cell count is a quantitative variable so this is a test for a difference in means among the
three groups, which is Analysis of Variance (ANOVA) for difference in means.
2. With only two treatment groups, this could be analyzed using either a t-test for a difference of two means
or using Analysis of Variance (ANOVA) for a difference in means.
3. D.20 Whether or not a person develops AIDS is categorical, so this is a test between two categorical
variables. A chi-square test for association is most appropriate.
4. With only two treatment groups, this could be analyzed using either a z-test for a difference of proportions
or a chi-square test for association.
5. D.26 The time it takes for a case to go to trial is a quantitative variable and this is a test for a difference in
means between the seven groups, which is Analysis of Variance (ANOVA) for difference in means.
6. D.27 Whether or not a case gets settled out of court is categorical, and the county is the other categorical
variable, so this is a test between two categorical variables. A chi-square test for association is most
appropriate.
7. Moore 27.43
a. This is a comparison of three means, so it requires ANOVA. The explanatory variable (high
money, low money, control) is categorical, and the response (number of pencils) is quantitative.
b. This is similar to the previous example, except that the explanatory variable (amount of Monopoly
money) is quantitative, so this has a quantitative explanatory and a quantitative response variable.
A regression test for slope or correlation is appropriate.
c. This is a comparison of three proportions, so it requires a chi-square test of association.
8. B.34 (a) Sample B. The sample mean in B is around 45, while the sample mean in A is around 47 and the
spread and sample size appear similar in the two samples. (b) Sample B. Both samples appear to have a
mean near 47, but the variability is smaller in sample B so we can be more sure the mean is below 50.
Also, sample B has only one value above 50, while sample A has 7 values above 50. (c) Neither. Both
samples appear to have means above 50, so neither would give evidence that the population mean is less
than 50.
9. B.35 (a) Sample A. The sample mean in A is around 43, while the sample mean in B is around 47. Sample
sizes and variability are similar for both samples. (b) Sample B. Both samples appear to have a mean near
46, but the variability is smaller in sample B so we can be more sure the mean is below 50. Also, sample
B has few values above 50, while sample A has at least 25% of its values above 50 (since Q3 > 50). (c)
Sample A. Both samples appear to have about the same mean and median (near 45) and similar
variability, but sample A is based on a much larger sample size, so it would be more unusual to see that
many values below 50 if H0 : μ = 50 were true.
10.
a. Dataset B shows stronger evidence, because the difference in means (variation between groups) is
about the same as that in Dataset A but the spread (variation within groups) for each sample is
much smaller.
b. Dataset B shows stronger evidence, because the difference in means (variation between groups) is
about the same as that in Dataset A but the spread (variation within groups) for each sample is
much smaller.
c. Dataset A shows stronger evidence, because the difference in means (variation between groups) is
much greater than that in Dataset B but the spread (variation within groups) for each sample is
about the same as in Dataset A.
11.
a.
Page 2 of 6
b.
c. (Number of rows – 1)•(Number of columns – 1) = 2•3 = 6
d. Conditions are not met. All expected counts are at least 5.
e. The χ2 test statistic is 14.8. This does not need to be calculated from scratch. It is just the sum of all
contributions to χ2.
f. P-value is approximately 0.02. (or between 0.01 and 0.04)
g. There is strong evidence of an association between the condition used on the dogs and the number
of errors. In particular, dogs in the nonsocial condition have zero errors much more than expected,
and dogs in the social-communicative condition have zero errors much less than expected. Dogs in
the social-communicative condition tend to have more errors.
h. Yes. It is a randomized experiment (the explanatory variable is randomly assigned).
i. 8/12 = 66.7%
j.
(0.400,0.934)
k. H0: pNS,0 = pNC,0
Ha: pNS,0 > pNC,0
One-tail p-value is slightly greater than 0.10
There is weak evidence that a higher proportion of dogs handled in the nonsocial manner would
have no errors than of dogs handled in the noncommunicative manner.
12.
a. Yes. All expected counts are at least 5.
b.

Page 3 of 6
iv) We do not have statistically significant evidence to show that the proportion of rainy days is
different among these four cities.
c.
i) Yes. All expected counts are at least 5.
ii) df = 3
iii)
iv) This is a small p-value, so we reject the null hypothesis, indicating that there is a difference
in the proportion of rainy days among the four seasons, and it appears the rainy season (with
almost twice as many rainy days as expected) is the winter.
13.
a. D.45

player who makes 100 free throws in a season will have between 339.3 and 1112.7 points for
the season.
b. D.46

a player who makes 400 free throws in a season will have between 1224.4 and 2008.5 points
for the season.
14.
a. ŷ = 560.7 – 3.077x
b. The average number of calories consumed by a toddler at the dinner table decreases by 3 calories
for each additional minute spent at the table beyond 20 minutes.
c. With 95% confidence, an individual toddler who spends 35 minutes at the table would consume
between 402.5 and 503.4 calories.
d. With 95% confidence, the mean number of calories consumed by a toddler who spends 35 minutes
at the table between 441.8 and 464.1 calories.
e. (422.232, 452.904) is the CI for the mean and (386.074, 489.063) is the PI for an individual. This is
evident because the CI for the mean is always more narrow than the PI for an individual.
f. There is a negative association, of moderate strength.
g. No. This is an observational study (the explanatory variable is not randomly assigned).
h. r = -0.65 (any estimate ±0.2 is acceptable)
15.
a. H0: Mean depression score is the same for all three sleep classifications.
Ha: Mean depression score is different for at least one sleep classification.
Page 4 of 6
b. Yes. The dotplots show heavy skew, but all samples are large enough to overcome this shape
(n≥246). Largest standard deviation = 4.719 and smallest = 2.56. These do not differ by more than
a factor of 2.
c. df = 2 for Groups
d. df = 895 for Error
e. P-value = 0.000. If the graph were extended to include the test statistic it would look like this:
f. There is overwhelming evidence that depression varies by sleep classification. Those with poor
sleep are more likely to be depressed than those with borderline or optimal sleep. Those with
optimal sleep are least likely to be depressed.
i) With 95% confidence, students with optimal sleep score, on average, between 6.582 and
7.444 on the POMS depression sub-scale.

iii) Yes. The difference in depression rating between optimal sleepers and poor sleepers is great.
This could be of high value to a person who suffers from depression, especially if a causal
relationship can be established. This observational study only indicates that the two variables
are related. However, it still may suggest that sleep symptoms and depression symptoms
should be addressed together.
16.
a. The null hypothesis is that all the state average home prices are equal and the alternative is that at
least two states have different means.
b. We are comparing k = 4 states, so the numerator degrees of freedom is k – 1 = 3.
c. The overall sample size is 120 homes, so the denominator degrees of freedom is n – k = 116
d. The underlying conditions to use an F-distribution for ANOVA are not met. In the histogram of NY
home prices we see that the distribution is heavily skewed and certainly not normal, but since we
have n = 30 in each group this is not a serious problem. The condition that is violated is actually the
equality of standard deviations. The standard deviation for California (sca = 1112.3) is more then 6
times that of Pennsylvania (spa = 179.4).
17.
a. Letting μ denote mean height we have the hypotheses
b. Yes, the conditions for ANOVA are satisfied. In three of the four groups the sample sizes are
greater than 30. In the group with only nT = 20 (tenors), the boxplot looks approximately
symmetric with no outliers, so the assumption of normality is not violated. The condition of equal
variability is satisfied because the within group standard deviations within each of the groups are
Page 5 of 6
not that different; the highest standard deviation (sT = 3.22), is less than twice the lowest (sS =
1.87).
c. P-value = 0.000
d. The p-value ≈ 0 provides overwhelming evidence against the null hypothesis. Average height of
singers differs by voice. Those who sing bass tend to be tallest, and those who sing soprano tend to
be shortest.
18. B.68 (a) A Type I error means the restaurant chain concludes that the arsenic level in chickens from a
supplier is too high (above 80), when actually the mean for the supplier is not more than 80 ppb. They
would cancel their orders from that supplier when they don’t need to. (b) A Type II error means that the
restaurant chain concludes that the arsenic level is acceptable from a supplier, when actually the mean is
more than 80 ppb. They would continue to buy chicken from this supplier, even though the mean arsenic
level is too high.
19. B.73 (a) In a Type I error, we conclude that treated wipes prevent infection, when actually they don’t. (b)
In a Type II error, we conclude that treated wipes are not shown to be effective, when actually they help
prevent infections. (c) A smaller significance level means we need more evidence to reject H0. We would
want a smaller significance level in the second situation (harmful side effects) so it has to be very clear
that the treated wipes help prevent infection, since we don’t want to put people at risk for side effects if
the benefit isn’t definite. (d) The p-value (0.32) is not small, so we do not reject H0. The study does not
provide sufficient evidence to show that treated wipes are more effective at reducing the proportion of
infected babies than sterile wipes. (e) Not necessarily. The results of the test are inconclusive when the pvalue is not small. Either H0 or Ha could still be valid, so the treated wipes might help prevent infections
and the study just didn’t accumulate enough evidence to verify it.
20. (a) 4.133 Type I error (releasing a drug that is really not more effective) when there are serious side
effects should be avoided, so it makes sense to use a small significance level such as α = 0.01. (b) 4.136
Type I error (suing the company when they are not lying) is quite serious so it makes sense to use a small
significance level such as α = 0.01. (c) 4.138 The company would prefer a large significance level, such
α = 0.10, which means they are more likely to find enough evidence to show that the drug works better.
Consumers would prefer a small significance level, such as α = 0.01, so they can be very sure that the
drug works better before paying the higher cost.
21. The p-value supports very strong evidence of some improvement but does not give any indication of the
magnitude. An average of 4.2 points for a test whose scores are in the hundreds is quite modest. If the
online prep course involves a substantial cost or time investment, this improvement probably does not
justify it. Although there is very strong statistical significance, we see very little practical significance.
22.
a. The randomization distibution would be centered at the null mean μ0 = 160
b. A t-distribution with 9 degrees of freedom would be appropriate.
c. The bootstrap distribution would be centered at = 135.
d. The bootstrap distribution would still show some skew, because n=10 is not sufficient for the
Central Limit Theorem to apply for a heavily skewed population.
23.
Page 6 of 6
a. The randomization distribution for should be centered at the null parameter, which is
zero.
b. See graph below. We draw a bell-shaped curve, centered at 0, and roughly locate the original
sample statistic, = 0.269, so that the area in the right tail is only about 0.02.
c. The bootstrap distribution would be centered at the sample statistic = 0.269.
24. B.15
a. The distribution on the left is more concentrated at the center, and therefore belongs with the larger
sample size (n=1000). Therefore, the distribution on the right belongs with n=100.
b. Both distributions are centered at the population parameter p=0.05.
c. The proportions for samples of size n = 100 go from about 0 to 0.12. The proportions for samples
of size n = 1000 go from about 0.025 to 0.07.
d. The standard error for samples of size n = 100 is about 0.02 (since it appears that about 95% of the
data are between 0.01 and 0.09.) The standard error for samples of size n = 1000 is about 0.005
(since it appears that about 95% of the data are between 0.04 and 0.06.)
e. A sample proportion of 0.08 is relatively likely from a sample of 100, but extremely unlikely with a
sample size of 1,000.

The post Elementary Statistics appeared first on My Assignment Online.

Expert Help With This Assignment — On Your Terms