Math 338 Summer
Lab Activity 7
Daniel Jordan
Application1:
A major car manufacturer wants to test a new engine to determine if it meets new air pollution
standards. The mean emission µ of all engines of this type must be approximately 20 parts per
million of carbon. If it is higher than that, they will have to redesign parts of the engine. Ten engines
are manufactured for testing purposes and the emission level of each is determined. Based on data
collected over the years from a variety of engines, it seems reasonable to assume that emission
levels are roughly normally distributed with σ= 3.
i.
What are the appropriate null and alternative hypotheses?
H0: µ = 20; Mean engine emission is 20 parts per million of carbon.
Ha: µ > 20; Mean engine emission is greater than 20 parts per million of carbon.
ii.
The data result in an average of 22 parts per million. What is the value of the test
statistic?
Zobserved = 0.67
iii.
What is the value of the P-value?
0.2524925
Application2: #6.71 Textbook.
Use R to perform test of hypothesis
6.71Are the mpg measurements similar? Refer to Exercise 6.26 (page 371). In addition to the
computer computing mpg, the driver also recorded the mpg by dividing the miles driven by the
number of gallons at each fill-up. The following data are the differences between the computer’s
and the driver’s calculations for that random sample of 20 records. The driver wants to determine if
these calculations are different. Assume the standard deviation of a difference to be σ = 3.0.
(a) State the appropriate H0 and Ha to test this suspicion.
H0: µ1 = µ2; The mean MPG calculations are equal
Ha: µ1 ≠ µ2; The mean MPG calculations are different
(b) Carry out the test. Give the P-value, and then interpret the result in plain language.
Since the p-value obtained from the data is 0.0175, which is less than 0.05, we can strongly reject the
null hypothesis of the mean gas mileage being equal. This tells us that the driver’s calculations differ
from the computer’s calculations.
Application 3
We wish to see if the dial indicating the oven temperature for a certain model oven is properly
calibrated. Four ovens of this model are selected at random. The dial on each is set to 300° F, and
after one hour, the actual temperature of each is measured. The temperatures measured are 305°,
310°, 300°, and 305°. Assume that the distribution of the actual temperatures for this model when
the dial is set to 300° is normal. To test if the dial is properly calibrated, we will test the following
hypotheses: H0: µ = 300 versus Ha: µ ≠ 300.
i.
Based on the R output, what is the value of the one-sample t statistic?
The value of the sample t statistic is 2.4495
ii.
Are the data statistically significant at the 5% significance level? interpret the
result in plain language.
Since the t test returned a p-value of 0.09172 which is greater than 0.05, we cannot reject the
hypothesis. And since Z0.05/2 = 1.645 is less than the t-value of 2.4495, we cannot reject the
hypothesis at a significance level of 5%. This tells us that the true mean temperature does not differ
from 300°
Application 4
A lab scientist is interested in whether lab rats that grow up alone grow as large as lab rats that
grow up with other rats around them to play with. He randomly selects ten young rats with
approximately the same age and size: Five of these will spend the next 4 months by themselves and
the other five rats will each have three other rats to play with during that same time. After 4
months, the scientist measures the abdomen circumference of all the rats (in mm). The results are
shown below.
Alone group (#1)
110
123
113
103
120
Play group (#2)
119
125
131
128
136
It seems reasonable to assume the variances in both populations are equal.
i.
What are the hypotheses to test whether the rats grow up to be equally large or
whether the playing rats grow up larger?
H0: µ1 = µ2; The mean abdomen difference is equal
Ha: µ1 ≠ µ2; The mean abdomen difference is not equal
ii.
Under the equal variance assumption, what are the appropriate degrees of freedom
for this test?
DF = 8
iii.
What is the value of the test statistic? What can we say about the value of the P-value?
Conduct the test and interpret the result in plain language.
t = -3.0638, p = 0.01549. Since p < 0.05, we can strongly reject the hypothesis, and conclude
that there is a difference in mean abdomen size of rats that grow up with a group rather
than alone.
iv.
Perform iii under the assumption of Unequal variance assumption.
t = -3.0638, p = 0.0164. Since p is still less than 0.05, we can again strongly reject the
hypothesis, and conclude that there is a difference in mean abdomen size of rats that grow
up with a group rather than alone.
Application 5
Sixteen people volunteered to be part of an experiment. All sixteen people were white, between the
ages of 25 and 35, and were supplied with nice clothes. Eight of the people were male and eight
were female. The question of interest in this experiment was whether females receive faster service
at restaurants than males. Each of the eight male participants was randomly assigned a restaurant,
and each of the eight females was randomly assigned to one of these same eight restaurants. One
Friday night, all sixteen people went out to eat, each one alone. The male and female assigned to the
same restaurant would arrive within 5 minutes of each other, with the order determined by flipping
a coin (male first or female first). Each person then ordered a similar drink and a similar meal. The
time (in minutes) until the food arrived at the table was recorded. They are shown below.
Restaurant
1
2
3
4
5
6
7
8
Male
22
14
16
26
18
13
9
27
Female
25
12
13
21
21
14
9
16
The hypotheses being tested are H0: µ= 0, Ha: µ > 0, where µ represents the mean of the
differences in service time (male – female). Assume that the differences are normally
distributed.
i.
What is the value of the t test statistic
t = 0.5888
ii.
What can you say about the value of the P-value? Conduct the test and interpret the
result in plain language.
Since p=0.2827 which is greater than 0.10, we cannot reject the null hypothesis. Thus,
we conclude that the difference in means between male and female service times are
equal.
Application 6:
Ten couples are participating in a small study on cholesterol. Neither the man nor the woman in
each couple is known to have any problems with high cholesterol. The researcher conducting the
study wishes to use the Matched Pairs to determine if there is evidence that the cholesterol level
for the husband tends to be higher than the cholesterol level for the wife. The cholesterol
measurements for the ten couples are given below
Couple
1
2
3
4
5
6
7
8
9
10
Husband’s cholesterol 224
310
266
332
244
178
280
276
242
260
Wife’s cholesterol
270
288
296
270
180
268
244
210
236
200
i.
What are the hypotheses the researcher wishes to test? What is the value of t test
statistic?
H0: µ1 = µ2; The husband’s mean cholesterol is equal to the wife’s mean cholesterol level
Ha: µ1 ≠ µ2; The husband’s mean cholesterol is greater than the wife’s mean cholesterol
level
t = 1.9822
ii.
What is the (approximate) value of the P-value? Conduct the test and interpret the
result in plain language.
p = 0.03939
Since 0.03939 < 0.05 we strongly reject the null hypothesis. With this, we can conclude that
husbands have a higher mean cholesterol level than their wives.