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Introduction to the Practice of Statistics
Sixth Edition
Moore, McCabe
Section 6.1 Homework Answers
6.10 Margin of error and the confidence interval.
A study based on a sample of size 25 reported a mean of 93 with a margin of error of 11 for 95% confidence.
(a) Give the 95% confidence interval.
93 ± 11 produces the interval (93 – 11, 93 + 11); (82, 104)
(b) If you wanted 99% confidence for the same study, would your margin of error be greater than equal
to, or less than 11? Explain your answer.
For a 99% confidence interval the margin of error would be larger since Z* for a 99% confidence
interval is larger than for a 95% confidence interval.
6.12 Changing the confidence level. A study with 36 observations had a mean of 70. Assume that the
standard deviation is 12. Make a diagram similar to Figure 6.6 (page 364) that illustrates the effect of the
confidence level on the width of the interval. Use 80%, 90%, 95%, and 99%. Summarize what the diagram
shows.
Z* for 80% is 1.282
Z* for 90% is 1.645
Z* for 95% is 1.960
Z* for 99% is 2.576
12
=2
36
36 ± 1.282(2) (33.436, 38.564)
σx =
36 ± 1.645(2) (32.71, 39.29)
36 ± 1.96(2) (32.08, 39.92)
36 ± 2.576(2) (30.848, 41.152)
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6.25 The mean number of calories consumed by women in the United States who are 19 to 30 years of age
is x = 1791 calories per day. The standard deviation is 31 calories.5 You will study a sample of 200 women
in this age range, and one of the variables you will collect is calories consumed per day.
(a) What is the standard deviation of the sample mean x ?
31

= 2.19 The purpose of this question is to make sure you understand what x =
really
200
n
means. Study how this question was phrased.
(b) The 68-95-99.7 rule says that the probability is about 0.95 that x is within
__2(2.19) = 4.38__ calories of the population mean μ. Fill in the blank.
The purpose of this question is to tie in the visual that we had in section 1.3, when the normal distribution
was first introduced along with the 68-95-99.7 rule. We know now that the numbers we associated with 6895-99.7 are just approximations so it would be easier to remember.
(c) About 95% of all samples will capture the true mean of calories consumed per day in the interval x plus
or minus 2(2.19) = 4.38 calories/day.
If you use the actual multiplier of 1.96 then the answer is
1.96(2.19) 4.29 cal/day.
The difference between (b) and (c) concerns what was discussed in class with respect how a confidence
interval works:
(b) There is approximately a 95% chance that x is in the interval ( - 4.38,  + 4.38).
(c) There is approximately a 95% chance that  is in the interval ( x - 4.38, x + 4.38).
6.26 Fuel Efficiency. Computers in some vehicles calculate various quantities related to performance. One
of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (MPG). For one vehicle
equipped in this way, the MPG were recorded each time the gas tank was filled, and the computer was then
reset.6
Here are the mpg values for a random sample of 20 of these records.
41.0
50.7
36.6
37.3 34.3
43.2 44.6
48.4
46.4
46.8
45.0 48.0 43.2 47.7
39.2 37.3
43.5
42.2
44.3
43.3
Suppose that the standard deviation is known to be  = 3.5 mpg.
 3.5 
(a) What is σ x , the standard deviation of x ? σ x = 
 = 0.7826
 20 
(b) Give a 95% confidence interval for , the mean mpg for this vehicle x = 43.15 mpg
43.15  1.96 (0.7826)
(41.62, 44.68)
6.32 Accuracy of a laboratory scale. To assess the accuracy of a laboratory scale, a standard weight known
to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean
(this mean is 10 grams if the scale has no bias). The standard deviation of the scale readings is known to be
0.0002 gram.
(a)The weight is measured five times- The mean result is 10.0023 grams. Give a 98% confidence interval for
the mean of repeated measurements of the weight.
0.0002
5
(10.0021, 10.0025)
10.00232.326
(b) How many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence?
2
 2.326(0.0002) 

 = 21.64. We need a sample of 22 values.
0.0001

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