1) For the following situations, indicate which type of hypothesis test would be used.
a. Children who have poor social skills are selected to participate in social skills
training. The researcher observes the children on the playground both before and
after the social skills training. For each of the 35 children, she records the number of
positive peer interactions. She then analyzes the mean before-after difference.
_____ one-sample t-test
__X___ matched pairs t-test
_____ two independent samples t-test
_____ pooled two sample t-test
b. A researcher randomly assigns 80 high school seniors to either an experimental group
where they learn about time management or to a control group where they discuss
current events. The researcher then gathers information, for all students, on
homework completion. The researcher wants to find out if the time management
session is effective for improving homework completion among high school seniors.
_____ one-sample t-test
_____ matched pairs t-test
___X__ two independent samples t-test
_____ pooled two sample t-test
c. A researcher want to know if chocolate affects your memory. The researcher find 40
pairs of twins, and randomly selects one twin to eat chocolate and the other twin does
not each chocolate. Then all 80 people are given a memory test. The researcher
records the score for each person.
_____ one-sample t-test
__X___ matched pairs t-test
_____ two independent samples t-test
__ ___ pooled two sample t-test
d. A researcher reads that the average time that adults in the United States spend
watching TV is 28 hours per week. The researcher randomly selects a sample of 40
college students. The researcher wants to find out if the students also watch TV an
average of 28 hours per week.
__X___ one-sample t-test
_____ matched pairs t-test
_____ two independent samples t-test
_____ pooled two sample t-test
e. College women who are being treated for depression at the campus counseling center
were examined for initial depression and then assigned to a drug treatment condition
or a placebo control condition. It is assumed the standard deviation is not affected by
the drug. At the end of the treatment period, the women take a depression test and
their test scores are analyzed to determine if the drug is helpful.
_____ one-sample t-test
_____ matched pairs t-test
_____ two independent samples t-test
___X__ pooled two sample t-test
2) A study timed left-handed and right-handed people to see how long it took them to hit a
buzzer. The data is shown below:
32 Right handed people had an average of 1.4 seconds and a standard deviation of 0.8 seconds
47 Left handed people had an average of 1.1 seconds and a standard deviation of 0.3 seconds
Test with 10% significance if right handed people are slower than left handed people.
H0: muR≤muL
HA: muR>muL
alpha = 0.10 (given in the problem – they can’t choose differently)
t = (1.4-1.1)/sqrt(.8^2/32+.3^2/47) = 2.0265
0.025<p-value<0.05
Reject
Our data shows right handed people are slower
3) A study attached microphones to people to count the words they spoke each day.
It is believed the standard deviations for both men and women should be equal.
The average for the 2 men was 6007 words with a standard deviation of 812 words.
The average for the 3 women was 7505 words with a standard deviation of 214 words.
Assuming the distribution of words is normal, test whether there is a difference in the number
of words spoken between men and women
H0: mu1=mu2
HA: mu1 ne mu2
Alpha=0.05
S=sqrt((1*812^2+2*214^2)/(2+3-2)) = 500
T=(6007-7505)/sqrt(500^2/2+500^2/3) = -3.28
2*0.02<p-value<2*0.025
0.04<p-value<0.05
Reject
Our evidence does show men and women use a different number of words on average
4) They say rock classical music is better for studying than rock music. To investigate the
difference a sample of 100 people were chosen to listen to a type of music while taking an IQ
test. Half listened to rock music, while the other half listened to classical music. The results
are shown below:
Classical Music group: 50 tests, average=104.3, standard deviation=9.4
Rock Music group: 50 tests, average=102.1, standard deviation=7.4
Find a 99% confidence interval for the difference of the two groups on average
(104.3-102.1)+-2.704*sqrt(9.4^2/50+7.4^2/50)=(-2.375, 6.775)
5) A study measured how long men and women stared at the new 144” HD HGTV. Of the 47
men the average was 12.4 minutes with a standard deviation of 5.3 minutes. Of the 27
women the average was 6.4 minutes with a standard deviation of 1.2 minutes. The average
difference was 6 minutes and the standard deviation of the differences was 0.68. Find a 92%
confidence interval for the average difference between the two genders.
Cannot do this problem – there are not enough women sampled
6) The distribution of wait times for a movie to load on Hulu is normally distributed with a
standard deviation of 2.4 seconds. What changes the mean is how long it takes Windows to
give the Hulu player priority in the process queue. I want to know the average on my old
computer, so I randomly sample it 4 times, and I get an average of 480 seconds. Find a 96%
confidence interval for the average time my old computer takes to play a movie on Hulu.
t=3.482 (475.82, 484.18)
7) Michael is planning to advertise by putting magnetic ads on cars, and he has an option of
putting the ads on white cars or red cars. To find out he convinces his friends to randomly
paint their cars either red or white (the order of which color was used first was chosen
randomly). Each car is parked in the Taco Bell parking lot and Michael notes how many
people stop to check out the car. Then the next day the car is painted the different color, and
parked in the same spot.
Michael assumes that the number of people who check out an ad on a car is normally distributed,
and he collected the data below. Show all 7 steps for the hypothesis test to determine if the
color of a car has an effect on the number of people who look at the car.
Red
Justin’s Car
Kenneth’s Car
Ryan’s Car
Omer’s Car
Sarah’s Car
Shawn’s Car
Variance of cars when painted red: 6.17 people2
Variance of cars when painted white: 10.27 people2
Unpooled sum of variances: 16.44 people2
7
5
8
7
12
10
White
10
8
7
12
11
16
Pooled variance of cars: 8.22 people2
Matched pairs variance: 8.70 people2
Variational Delta: 4.1 people2
First it is important to notice that this is a matched pairs t-test.
The data is paired by each individual’s car. This is not ANOVA because pairs of data are linked,
not independent. This is not regression because the colors red and white are not numerical. This
is not a chi-squared goodness of fit test because we are not testing whether Michael’s friends
tend to have red cars or white cars. In fact the only reason we care about who’s car it was is to
emphasize that the data is paired along each row.
The mean is -2.5 (red-white), variance 8.7, sample size is 6 (that’s all we need)
H0: μred=μwhite
HA: μred≠μwhite
α=0.05
 2.5  0
t 61 
 2.08
8.7
6
0.025<tail value<0.05
Double because this is a two-tailed test
0.05<p-value<0.1
Since p-value>.05, Fail to reject
Conclude that there is not enough evidence to say that the color of the car affects the number of
people who look at the car.
8) An engineer believes the average weight of a tree is 6000 pounds. He feels it’s safe to
assume the weights are normally distributed. He weighs three randomly selected trees and
records their weight:
6000 pounds
13000 pounds
51500 pounds
Test whether the engineer is correct.
H0:mu=6000
HA:mu ne 6000
Alpha = 0.05 (or something reasonable)
Xbar = 23500
Sd=sqrt((6000-23500)^2+(13000-23500)^2+(51500-23500)^2)/(3-1))
Sd = 24500
Df=2
T=(23500-6000)/(24500/sqrt(3))=1.24
2*.15< p-value <2*0.2
.3 < p-value < .4
Fail to Reject
Our data does not show the rancher is incorrect
9) An investigator wishes to determine whether alcohol consumption causes deterioration in the
performance of automobile drivers. Before the driving test, subjects drank a glass of orange
juice which, in the case of the treatment group, is laced with two ounces of vodka.
Performance is measured by the number of errors made on a driving simulator. 120
volunteer subjects are randomly assigned, in equal numbers, to the two groups. It is believed
that the variance in both groups will be the same. For subjects in the treatment group, the
mean number of errors (𝑥̅1 ) equals 26.4 and the standard deviation (𝑠1 ) equals 13.99. For
subjects in the control group, the mean number of errors (𝑥̅2 ) equals 18.6, and the standard
deviation (𝑠2 ) equals 12.15. Find a 95% confidence interval for the difference in errors for
the alcohol group compared to the control group.
sp 
60  113.99 2  60  112.15 2
60  60  2
𝟐
=13.1
𝟐
𝟏𝟑.𝟏
𝟏𝟑.𝟏
𝟐𝟔. 𝟒 − 𝟏𝟖. 𝟔 ± 𝟏. 𝟗𝟖𝟒√ 𝟔𝟎 + 𝟔𝟎
=(3.05, 12.55)
10) A database manager wants to know if plugging in a speaker slows down a computer. She
randomly selects 7 different computer companies and gets two computers from each
company. For each company she has one computer where speakers are plugged in, and one
where they are not. Then all 14 computers calculate the ten billionth digit of pi. Assume
computer times are normally distributed and that the speakers do not affect the standard
deviation. Given the data below calculate a 99.8% confidence interval for the difference in
average time.
Computers without speakers
Computers with speakers
Computers: 7
Computers: 7
Mean time: 850 minutes
Mean time: 910 minutes
Standard deviation: 25 minutes
Standard deviation: 45 minutes
The standard deviation of the differences: 21 minutes
Pooled Standard deviation: 36.4 minutes
THIS one is the matched pairs example (each pair is of the same company). Also a t=5.208
(-101.27, -18.66)
11) 40 volunteers were selected to take part in a “Twinkie diet” experiment. The subjects
were weighed first. Of the two who were the lightest weight one was chosen to be
control and the other “Twinkie”. Then of the next two highest weights one was control,
the other “Twinkie”. This pattern continued until half were in the Twinkie group and the
other half were the control group. The data is shown below.
Twinkie group:
The twenty people averaged 40 pounds gained with a standard deviation of 22 pounds
Control group:
The twenty people averaged 8 pounds gained with a standard deviation of 10 pounds
Matched pairs standard deviation: 6 pounds
Pooled standard deviation: 17 pounds
Find a 90% confidence interval for the difference in average pounds gained.
Not a large enough sample size to do