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Homework
Hypothesis Testing
1. Shell Oil claims that the average gasoline price of their brand is less than $1.25
per gallon. In order to test their claim, we randomly selected a sample of 49 of
their gas stations and determined that the average price per gallon of the stations
in the sample is $1.20. Furthermore, we assume that the standard deviation of the
population is $.14 from other studies. Using an alpha of .05, test the company’s
claim. State a conclusion that would be understandable to anyone.
Next, find the p-value and explain its meaning.
2. The manager of a local grocery store believes that the store’s average daily sales
are more than $8,000 per day. To test his belief, a sample of 64 days of sales was
selected and it was found that the average sale was $8,250 with a sample standard
deviation of $1,200. Test the manager’s belief using an alpha of .01. State a
conclusion that would be understandable to anyone.
Next, find the p-value and explain its meaning.
3. A lathe is set to cut bars of steel with a length of 6 centimeters. The lathe is
considered to be in perfect adjustment if the average length of the bars it cuts is 6
centimeters. A sample of 25 bars is selected randomly, and their lengths are
measured. It is determined that the average length of the bars in the sample is 6.1
centimeters with a standard deviation of .2 centimeters. At an alpha value of .05,
test to see if the lathe is in perfect adjustment. State a conclusion that would be
understandable to anyone.
Next, find the p-value and explain its meaning.
At an alpha value of .01, what would our conclusion be and why?
4. More professional women than ever before are foregoing motherhood because of
the time constraints of their careers. Yet, many women still manage to find time
to climb the corporate ladder and set time aside to have children. A survey of 187
attendees at Fortune Magazine’s Most Powerful Women in Business summit in
March 2002 found that 133 had at least one child. Assume the group of 187
women is a random sample from the population of all successful women
executives.
a. What is the sample proportion of successful women executives who have
children?
b. At the .05 level of significance, can you state that more than half of all
successful women executives have children?
c. At the .05 level of significance, can you state that more than two-thirds of
all successful women executives have children?
d. Do you think the random sample assumption is valid? Explain.
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Homework
Hypothesis Testing Using P-Value’s
1.
In the summer of 2001, Congress approved a federal budget containing several
provisions for tax reductions. The claim was that it would save the average
taxpayer $800. A sample of 500 taxpayers showed a mean reduction in taxes of
$785.10 with a standard deviation of $277.70. Test the hypothesis at the 5% alpha
level using the p-value. Interpret the p-value and comment on its meaning.
2.
Ben and Jerry sell ice cream from a pushcart in NYC Central Park. Ben tells
Jerry they sell an average of at least 15 pounds of vanilla when the temperature
exceeds 80 degrees. If 20 days of at least 80 degree weather reveals an average of
13.9 pounds with a standard deviation of 2.3 pounds, who is correct at the alpha
level of .05 using the p-value. Interpret the p-value and comment on its meaning.
3. Use Excel to get P-values using the data file MOISTURE. Boston and Vermont
make asphalt shingles. Builders (the customers for the shingles) may feel they
have purchased a product lacking in quality if they find moisture (measured in
pounds per hundred square feet) and wet shingles inside the packaging. Excess
moisture can cause appearance and construction problems. The manufacturers
monitor the amount of moisture present by taking samples of the singles received.
The file MOISTURE contains moisture levels for a sample from each
manufacturer. Each manufacturer would like to show that the moisture levels
for their product is less than .35.
Use the t methodology even though the sample sizes are slightly greater than 30.
a. What are the Null and Alternative Hypothesis?
b. Use Excel to determine the p-value for each sample.
c. What conclusions can you make?
Use Excel to complete the following steps:
 Find the mean of each sample.
 Find the standard deviation of each sample.
 Find n and the degrees of freedom for each sample.
 Calculate the test statistic t value for each sample.
 Find the P-Value for each sample. Use the Excel function
TDIST(ABS(t-test value), d.f., 1 tailed test)
 What conclusions would you make at the .05 LOS?
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Hypothesis Testing – Marketing Survey Example
An advertising firm has a client that is interested in buying an expensive
television commercial for air during the 2007 super bowl. The client will buy the
commercial if the ad firm can provide strong evidence that their commercial
provokes a sense of excitement in the viewer.
The commercial has been filmed and shown to two groups of likely super bowl
viewers. After viewing the commercial, each person completed a questionnaire
asking them about their feelings of excitement regarding the product that was
shown in the commercial. Feelings of excitement was measured on a 1 to 10
point scale with 10 being very excited about the product and 0 being not excited
at all. The ratings for each person in each group are contained in the
spreadsheet files labeled SampleA and SampleB.
The ad firm claims their ad provokes an excitement level rating of more than 6 on
a 10-point scale. If this claim can be substantiated, the client will proceed with
the purchase.
1. Perform two hypothesis tests (for groups A and B) at the 95% level (the same
test but for each sample group). Clearly show all the steps performed. State a
short conclusion and then an explanation of your conclusion relevant to the
context of this scenario.
2. For sample group B, how confident could you be in your conclusion? Explain
how you arrived at this result.
Helpful formula:
=average(cell range)
=stdev(cell range)
mean
sample standard deviation
=tdist(t test statistic value, df, number of tails)
this gives the area under the t-distribution in a tail(s), to obtain the p-value
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Module 4, Homework
Excel for p-values
Linear Regression, Part 1
Excel p-values (first use the table to estimate, then use Excel):
1.
2.
3.
4.
For a Type II hypothesis test with t*=1 and n=15 determine the p-value.
For a Type III hypothesis test with t*=1.8 and n=21 determine the p-value.
For a Type I hypothesis test with Z*=-2.16 determine the p-value.
For a Type III hypothesis test with Z*=-1.11 determine the p-value.
Linear Regression Problem:
The marketing manager of a large supermarket chain would like to determine the effect
of shelf space on sales of pet food. A random sample of 13 stores is selected and the
results are shown in the Excel data file PETFOOD.
a. Set up a scatter diagram to see if the potential exists for a linear relationship
between sales and shelf space.
Highlight from A1:B14
Click on the Chart Wizard icon
Select XY(Scatter)
Select next and get to the labeling window
Properly label the X and Y axis
Select Chart, Add Trendline
b. Perform an Excel Linear Regression analysis to determine the regression
coefficients.
Select Tools, Data Analysis, Regression
For the Y-Range, enter B1:B14
For the X-Range, enter A1:A14
Check the Labels box
In the new worksheet ply box type in: LR Output
c. Interpret the meaning of the slope.
d. Predict the average weekly sales of pet food for a store with 17 feet of shelf space.
e. Determine the error between actual sales and estimated sales for your answer in d.
f. Interpret the coefficient of determination for this situation.
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ANSWERS
Hypothesis Testing HW
1. Type I test, Zcrit=-1.65, Ztest=-2.5, reject H0 and conclude the mean gas prices
are less than $1.25 at the .05 LOS (we are 95% confident). P-value=.0062, very
small, so it is very unlikely we have incorrectly rejected H0 (we are very likely
right, 99.38% confident, in concluding the mean gas price is less than $1.25).
2. Type II test, t-crit=2.39, t*=1.67, cannot reject H0. The manager’s claim that
average daily sales exceed $8000 cannot be supported at the alpha=.01 LOS.
The p-value=.05, so we could be 95% confident in rejecting H0 and concluding
average daily store sales are greater than $8000. Thus, based on the mediocre pvalue, we can be somewhat confident, but not extremely confident that average
daily store sales are greater than $8000.
3. Type III t-test, t-crit=+/-2.064, t-test=2.5, reject H0, accept HA and conclude the
lathe is not in adjustment at the .05 LOS. P-value is between .02 and .01. We can
be up to 98% confident in our conclusion. At alpha=.01 we could not reject H0
and would conclude the lathe was in adjustment (mu=6).
4. a. Ps = .71
b. Z* = 5.7 > Z crit value of 1.645, reject H0 and conclude there is enough
evidence at the .05 LOS that more than half the women execs have children.
c. Z* = 1.29 < Z crit value of 1.645, do not reject H0 and conclude there is not
enough evidence at the .05 LOS that more than two-thirds of the women execs
have children.
d. The random sample assumption is not likely valid because we sampled the
“Fortune most powerful women” this is not the same as “successful women
execs.”
Hypothesis Testing – Marketing Survey
1. Group A: t*=1.3489, X-bar = 6.7, S = 2.84, CV= 1.699
pvalue=.0937
We cannot conclude with 95% confidence that the mean excitement rating
produced by the ad is greater than 6.
2. Group B: t*=2.16, X-bar = 6.7, S = 2.5255, CV= 1.671
pvalue=.0172
We can conclude with 95% confidence that the mean excitement rating produced
by the ad is greater than 6. (note: the bigger sample size helped us)
We could be 98.28% confident in our conclusion.
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Hypothesis Testing Using P-Values
1. H0: µ = 800
H1: µ ≠ 800
Z* = -1.2 < Zcrit value of +/- 1.96, do not reject H0.
P-value is .23 > α of .05, do not reject H0. If we were
to reject H0, we could only do so with 77% confidence.
There would be a 23% chance (this is large) of incorrectly
rejecting (rejecting H0 when it is true). Thus, we cannot
reject H0 with a great deal of confidence and we must
conclude taxpayers do receive about an $800 reduction on
average.
2. H0: µ ≥ 15
H1: µ < 15
t* = -2.14, the p-value is close to .025 (just smaller)
which is less than α of .05, so we reject H0 and conclude,
Ben is wrong, they are selling less 15 lbs. of vanilla when
the temperature exceeds 80 degrees. We make this
conclusion with at least 97.5% (1-.025) confidence (much
confidence).
3. H0: µ ≥ .35
H1: µ < .35
Mean for Boston sample of 36 is .3167, s=.1357.
Mean for Vermont sample of 31 is .2735, s=.137.
t* Boston = -1.47, p-value .075
t* Vermont = -3.1, p-value .002
At a .05 level of significance, we can
Vermont sample only (since its p-value
Thus, we conclude the Vermont shingles
level less than .35. We can make this
very large (1-.002) confidence level.
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reject H0 for the
is less than .05).
have a moisture
conclusion with a
Excel and P-Values
1.
2.
3.
4.
table: between .15 and .25, Excel .167
table: between .05 and .10, Excel .087
.0154
.267
Linear Regression 1 Pet Food
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.833982217
R Square
0.695526338
Adjusted R Square
0.667846914
Standard Error
0.305103313
Observations
13
69.55% of the variation in sales is explained by pet
food shelf space allocation.
ANOVA
df
Regression
Residual
Total
Intercept
Space-X
SS
MS
F
Significance F
1 2.339109 2.339109 25.12792 0.000395
11 1.023968 0.093088
12 3.363077
CoefficientsStandard Error t Stat
P-value Lower 95%Upper 95%
1.432883939 0.214896 6.667805 3.53E-05 0.959901 1.905867
0.077080891 0.015377 5.012776 0.000395 0.043237 0.110925
b0 = 1.43
b1 = .077
Y-hat = 1.43 * .077(Space-X)
b1 interpretation: For each 1 foot increase in shelf space, the expected increse
in weekly sales is .077 (hundred dollars) or $7.70.
predicted weekly sales for 17 feet of shelf space: $2.739 hundred dollars or $273.90.
actual weekly sales for 8 feet of shelf space: $3 hundred dollars or $300.
error: $26.10
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