Chapter 7 Section 1
Homework Set B
Data is available at CrunchIt! site: bcs.whfreeman.com/crunchit/ips6e
On the left side of the worksheet page you will see a links to chapters; click on chapter 7, then select appropriate
Exercise.
7.32 Food intake and weight gain. If we increase our food intake, we generally gain weight. Nutrition scientists can
calculate the amount of weight gain that would be associated with a given increase in calories. In one study, 16
nonobese adults, aged 25 to 36 years, were fed 1000 calories per day in excess of the calories needed to maintain a
stable body weight. The subjects maintained this diet for 8 weeks, so they consumed a total of 56,000 extra calories. 12
According to theory, 3500 extra calories will translate into a weight gain of 1 pound. Therefore, we expect each of these
subjects to gain 56,000/3500 = 16 pounds (Ib). Here are the weights before and after the 8-week period expressed in
kilograms (kg):
A) Describe the distribution using graphical methods. Is it appropriate to analyze these data using methods based on
Normal distributions?
Normal Quantile Plot
4
2
5
3
2
4
8
8
7
6
6
5
3
8
6
0
0
9
4
9
5
6
Differences
1
5
4
3
2
1
7
0
7
1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Expected Z-score
The stemplot seems to suggest there may be two distribution here, it has that bimodal look. Both curves however
seem fairly normal. The normal quantile plot suggests that the distribution is normal. Maybe that dip in the
stemplot is just an annomaly and the distribution is normal. Since there are no extreme outliers and our sample size
is 16 I feel safe using a t distribution for my calculations.
(a) For each subject, subtract the weight before from the weight after to determine the weight change.
id
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Wt Bef
55.7
54.9
59.6
62.3
74.2
75.6
70.7
53.3
73.3
63.4
68.1
73.7
91.7
55.9
61.7
57.8
Wt Aft
61.7
58.8
66
66.2
79
82.3
74.3
59.3
79.1
66
73.4
76.9
93.1
63
68.2
60.3
Difference
6
3.9
6.4
3.9
4.8
6.7
3.6
6
5.8
2.6
5.3
3.2
1.4
7.1
6.5
2.5
(b) Find the mean and the standard deviation for the weight change.
x diff = 4.73
s = 1.75
(c new) Theory suggest that the weight gain by the individuals should be 16lbs on average. Is there any evidence to
suggest that the average weight gain is different? Run a significance test, and use a 5% confidence level. Write down
the null and alternative hypothesis, the necessary calculations, the p-value, and the conclusion.
Ho:  = 16 lbs Ha:   16 lbs.
x diff = 4.73
s = 1.75



4.73 - 16 
P( x diff < 4.73 ) = P  t 

1.75 



16 

=P(t < -25.76) degrees of freedom 15.
0
p-value  2(0) = 0
The calculation is merely a perfunctory act, we can clearly see, given the sample mean value and standard
deviation that we are nowhere near 16lbs. The chance of seeing a weight as low as the one we just
encountered or even lower, when we assume  = 16lbs is nearly impossible to achieve. Thus, there is
overwhelming evidence to suggest that the mean weight gain is not 16lbs.
(c) Calculate the standard error and the margin of error for 95% confidence. Report the 95% confidence interval in a
sentence that explains the meaning of the 95%.
S.E. =
1.75
 0.4375
16
t* = 2.131 which I can get from Excel by typing =tinv(0.05, 15) or by looking at the t-table.
4.73  2.131(0.4375)
(3.79 lbs, 5.66 lbs)
(f) Write a short paragraph explaining your results.
The 95% confidence inteval produces a possible range of averages in which the true weight gain from the
population, subject to the eating regiment mentioned, may lie. We are 95% certain that the true weight gain is in
the calculated interval. The interval does not support the theory that the mean weight gain would be 16lbs.
7.34 Potential insurance fraud? Insurance adjusters are concerned about the high estimates they are receiving from
Jocko's Garage. To see if the estimates are unreasonably high, each of 10 damaged cars was taken to Jocko's and to
another garage and the estimates recorded. Here are the results:
Test the null hypothesis that there is no difference between the two garages. Be sure to specify the null and alternative
hypotheses, the test statistic with degrees of freedom, and the P-value. What do you conclude?
A) Describe the distribution using graphical methods. Is it appropriate to analyze these data using methods based on
Normal distributions?
Ho:  = 0 Ha:  > 0
Normal Quatile Plot
350
300
250
Differences
The normal quantile plot shows the data following a straight line
pattern that is indicative of a normal distribution. The data value
on the left is not extreme enough, but one value is acceptable,
meaning we can attribute it to chance that one value is off the
mark. Thus we have no evidence to suggest the data is not
normally distributed. I feel confident using a t-distribution for
my calculations.
200
150
100
50
0
-2
x diff = $115.0
s = $124.83



115 - 0 
P( x diff > 115 ) = P  t 

124.83 



10 

=P(t > 2.91) degrees of freedom 9.
 0.0087
-1.5
-1
-0.5
-50
0
0.5
-100
Expected Z-score
1
1.5
2
The p-value is 0.0087, the strength of the evidence is good that Jocko’s shop is charging more
on average than the other garage.
115.0  2.262
124.83
10
t* found by typing =tinv(.05, 9) or looking at table.
(25.71, 204.29)
The 95% confidence interval suggest that indeed Jockos garage is charging more on average than the other garage,
and we are 95% confident the true average is in the interval we calculated.
7.35 Fuel efficiency comparison t test. Refer to Exercise 7.24. In addition to the computer calculating mpg, the driver
also recorded the mpg by dividing the miles driven by the amount of gallons at fill-up. The driver wants to determine if
these calculations are different.
Problem 24 Computers in some vehicles calculate various quantities related to performance. One of these is the
fuel efficiency,..
A) Describe the distribution using graphical methods. Is it appropriate to analyze these data using methods based on
Normal distributions?
Normal Quantile Plot
10
8
Differences
6
4
2
0
-2.5
-2
-1.5
-1
-0.5
0
0.5
-2
-4
-6
Expected Z-score
1
1.5
2
2.5
(a) State the appropriate Ho and Ha.
Ho:  = 0 Ha:   0. We are going to asume there is no difference in the average difference between the two
measurements.
(b) Carry out the test. Give the P-value, and then interpret the result.
x diff = 2.7mpg
s = 2.80 mpg



2.7 - 0 
P( x diff > 2.7 ) = P  t 

2.80 



20 

=P(t > 4.31) degrees of freedom 19.
 0.00019
P-value = 2(0.00019)
= 0.00038
There is strong evidence that there is a difference between the computer and driver calculations.
(c) If the result is statistically significant at the 1% level calculate a 95% confidence interval and interpret what the
interval is trying to tell us.
2.7  2.093
2.80
20
degree of freedom is 19, t* is found by table or by typing =tinv(0.05,19)
(1.39, 4.01) We are 95% certain that the true difference between the computer and driver measurements is in the
calculated interval. Since the difference was calculated as computer – driver, the postive values suggest that the
drivers calculation is on average lower than the computers calcuation; the interval points somewhere between
1.39mpg to 4.01mpg.
7.44 Property damage due to tornadoes. Table 1.5 (page 25) gives the average property damage per year due to
tornadoes for each of the 50 states and Puerto Rico. It does not make sense to use the t procedures (or any other
statistical procedures) to give a 95% confidence interval for the mean property damage per year due to tornadoes in
the United States. Explain why not.