Test Statistic Excel functions Confidence Interval formula One population mean sigma known H0:  = value H1:  ><≠ value z = x - s / n =NORM.S.DIST( ) X ± Za /2 s n =NORM.S.INV(a/2) One population mean sigma unknown H0:  = value H1:  ><≠ value t = x - s / n =T.DIST( ) =T.DIST.RT( ) =T.DIST.2T( ) (n-1) degrees of freedom X ± ta /2 s n = T.INV.2T(an-1) One population proportions H0: p= value H1: p ><≠ value n/)p1(p pp̂ z - - = =NORM.S.DIST( ) (n*p and n*(1-p) are greater than 5) �̂� ± 𝑍 / �̂�(1 − �̂�) 𝑛 =NORM.S.INV(a/2) CHI-SQUARED TESTS Test statistic: 𝜒 = ∑ ( ) ; Expected value: 𝑒 = ( row total)( column total) Total sample size Degrees of freedom goodness of fit: (rows–1); Degrees of freedom independence: (rows-1)*(columns-1) ANOVA Source of Variation Sum of Squares Degrees of Freedom Mean Squares F-statistic Factor (between) SSB k-1 MSB=SSB/(k-1) F=MSB/MSW Error (within) SSW n-k MSW=SSW/(n-k) Total SST n-1 Source of Variation Sum of Squares Degrees of Freedom Mean Squares F-statistic Factor A SSA a-1 MSA=SSA/(a-1) F=MSA/MSW Factor B SSB b-1 MSB=SSB/(b-1) F=MSB/MSW Interaction SSAB (a-1)*(b-1) MSAB=SSAB/(a-1)*(b-1) F=MSAB/MSW Error (Within) SSW n-(a*b) MSW=SSW/(n-(a*b)) Total SST n-1 REGRESSION 𝑅 = Prediction Interval: DESCRIPTIVE: Mean: �̅� = ∑ Variance: 𝑠 = ∑ ( ̅) Q1 A random variable X is normally distributed with a mean of ten and a standard deviation of five. A sample of size n=4 was taken from this population. a. (1 pt) What is the probability that the sample mean is greater than twelve? b. (1 pt) Would your ability to answer the question change if you were told that X is not normally distributed? Why? Q2Table AHow Typical Students Spend Their TimeDaily ActivitiesHypothesized HoursKnown standard deviationSample mean (n=50)Sleeping8.727.8Leisure and Sports4.10.54.2Educational Activities3.332.4Working2.422.9Other2.30.73.1Traveling1.40.21.4Eating and Drinking10.71.2Grooming0.80.31Total2424h. An upper-tail test for the average hours spent grooming (use α=0.05) The data in Table A describes time use for an average weekday for full- time college students. The first column provides various types of activities, the second column provides the hypothesized number of hours a typical student spends on each activity (taken from the website of the American Bureau of Labor Statistics), and the third column provides the population standard deviation (σ) for each activity. To test whether these data are reflective of student time use at your school, you selected a sample of 50 of your classmates and asked them to provide data on their time use for an average weekday. The sample means you calculated for each activity are provided in the fourth column. Using the information provided in Table A, find the p-value for each of the tests posed in questions a.-h., using

1. Test Statistic Excel functions Confidence Interval formula
One population mean
sigma known
z = x -
s / n
=NORM.S.DIST( )
X ± Za /2
s
n
=NORM.S.INV(a/2)
One population mean
sigma unknown
t = x -
s / n

2. =T.DIST( )
=T.DIST.RT( ) =T.DIST.2T( )
(n-1) degrees of freedom
X ± ta /2
s
n
-1)
One population
proportions
H1: p ><≠ value
n/)p1(p
pp̂
z
-
-
= =NORM.S.DIST( )
(n*p and n*(1-p) are
greater than 5)
�̂� ± � /

3. �̂�(1 − �̂�)
�
=NORM.S.INV(a/2)
CHI-SQUARED TESTS
Test statistic: � = ∑
( )
; Expected value: � =
( row total)( column total)
Total sample size
Degrees of freedom goodness of fit: (rows–1); Degrees of
freedom independence: (rows-1)*(columns-1)
ANOVA
Source of Variation Sum of
Squares
Degrees of
Freedom
Mean Squares F-statistic

4. Factor (between) SSB k-1 MSB=SSB/(k-1) F=MSB/MSW
Error (within) SSW n-k MSW=SSW/(n-k)
Total SST n-1
Source of Variation Sum of
Squares
Degrees of
Freedom
Mean Squares F-statistic
Factor A SSA a-1 MSA=SSA/(a-1) F=MSA/MSW
Factor B SSB b-1 MSB=SSB/(b-1) F=MSB/MSW
Interaction SSAB (a-1)*(b-1) MSAB=SSAB/(a-1)*(b-1)
F=MSAB/MSW
Error (Within) SSW n-(a*b) MSW=SSW/(n-(a*b))
Total SST n-1
REGRESSION
� = Prediction Interval:
DESCRIPTIVE:
Mean: �̅� =
∑

5. Variance: � =
∑ ( ̅)
Q1
A random variable X is normally distributed with a mean of ten
and a
standard deviation of five. A sample of size n=4 was taken from
this
population.
a. (1 pt) What is the probability that the sample mean is greater
than twelve?
b. (1 pt) Would your ability to answer the question change if
you were told that X
is not normally distributed? Why?
Q2Table AHow Typical Students Spend Their TimeDaily
ActivitiesHypothesized HoursKnown standard deviationSample
mean (n=50)Sleeping8.727.8Leisure and
Sports4.10.54.2Educational
Activities3.332.4Working2.422.9Other2.30.73.1Traveling1.40.2
1.4Eating and Drinking10.71.2Grooming0.80.31Total2424h. An
upper-tail test for the average hours spent grooming (use
α=0.05)
The data in Table A describes time use for an average weekday
for full-
time college students. The first column provides various types
of activities,
the second column provides the hypothesized number of hours a
typical
student spends on each activity (taken from the website of the
American
Bureau of Labor Statistics), and the third column provides the
population

6. standard deviation (σ) for each activity. To test whether these
data are
reflective of student time use at your school, you selected a
sample of 50 of
your classmates and asked them to provide data on their time
use for an
average weekday. The sample means you calculated for each
activity are
provided in the fourth column. Using the information provided
in Table A, find the p-value for each of the tests posed in
questions a.-h., using
the provided value of α to draw your conclusion of whether or
not to reject
the null hypothesis.
a. (1 pt) A two-tail test for the average number of sleep hours
(use α=0.1)
b. (1 pt) An upper-tail test for the average hours spent on
leisure and sports (use α=0.1)
c. (1 pt) A lower-tail test for the average hours spent on
educational activities (use α=0.05)
d. (1 pt) A two-tail test for the average hours spent working
(use α=0.01)
e. (1 pt) An upper-tail test for the average hours spent on other
activities (use α=0.01)
f. (1 pt) A two-tail test for the average hours spent traveling
(use α=0.05)
g. (1 pt) An upper-tail test for the average hours spent eating
and drinking (use α=0.05)
h. (1 pt) An upper-tail test for the average hours spent grooming
(use α=0.05)
Q3
(2 pt) The United States government claims DUI arrests
average 22,096 per state
per year. A sample of size n=10 states finds the mean to be
44,002. If DUI
arrests are normally distributed with a standard deviation of

7. 28,584, can it
be concluded at the 1% significance level that DUI arrests are
higher than
the government claims?
Q4
Joe wants to know if he receives the same number of emails as
his co-
workers, who claim to get seventeen emails daily with a known
population
standard deviation of five emails. He takes a random sample of
twenty days
and records how many emails he receives. His sample average is
17.9
emails per day.
a. (1 pt) Construct a 95% interval for the mean number of daily
emails.
b. (1 pt) Can Joe conclude at the 5% significance level that he
gets the same
number of emails per day as claimed by his co-workers?