Download Actually, like the z-interval procedure, the t

440
CHAPTER 8
CONFIDENCE INTERVALS FOR ONE POPULATION MEAN
Actually, like the z-interval procedure, the t-interval procedure works reasonably well even when the variable under consideration is not normally distributed
and the sample size is small or moderate, provided the variable is not too far from
being normally distributed. In other words, the t-interval procedure is robust to
moderate violations of the normality assumption.
When considering the t-interval procedure, it is also important to watch for
outliers. The presence of outliers calls into question the normality assumption. And
even for large samples, outliers can sometimes unduly affect a t-interval because
the sample mean and sample standard deviation are not resistant to outliers.
Guidelines for when to use the t-interval procedure are the same as those
given for the z-interval procedure in Key Fact 8.1 on page 421. And remember,
always look at the data before applying the t-interval procedure to ensure that it is
reasonable to use it.
EXAMPLE
8.9
Illustrates Procedure 8.2
The U.S. Federal Bureau of Investigation (FBI) compiles data on robbery and
property crimes and publishes the information in Population-at-Risk Rates and
Selected Crime Indicators. A sample of last year’s pick-pocket offenses yields the
values lost shown in Table 8.5. Use the data to obtain a 95% confidence interval
for the mean value lost, µ, of all last year’s pick-pocket offenses.
TABLE 8.5
Value lost ($)
for a sample of
25 pick-pocket offenses
Solution
447
313
217
833
649
207
844
768
277
554
627
253
1064
805
570
430
397
26
653
223
883
214
587
549
443
Since the sample size, n H 25, is moderate, we first need to consider questions of
normality and outliers. (See the second bulleted item in Key Fact 8.1 on page 421.)
To do that, we constructed a normal probability plot for the data in Table 8.5, as
shown in Fig. 8.8.
The normal probability plot in Fig. 8.8 shows no outliers and falls roughly in
a straight line. Thus we can apply Procedure 8.2 to obtain the required confidence
interval.
Step 1
For a confidence level of 1 − α, use Table IV to find tα/2 with
df H n − 1, where n is the sample size.
We want a 95% confidence interval, so α H 1 − 0.95 H 0.05. Since n H 25,
df H 25 − 1 H 24. Consulting Table IV, we find that tα/2 H t0.05/2 H t0.025 H 2.064.
8.4
441
3
2
Normal score
FIGURE 8.8
Normal probability
plot of the value-lost
data in Table 8.5
CONFIDENCE INTERVALS FOR ONE POPULATION MEAN WHEN σ IS UNKNOWN
1
0
21
22
23
0
200 400 600 800 1000 1200
Value lost ($)
Step 2 The confidence interval for µ is from
s
s
x − tα/2 · √
to x + tα/2 · √ .
n
n
From Step 1, tα/2 H 2.064. Applying the usual formulas for x and s to the data
in Table 8.5, we find that x H 513.32 and s H 262.23. Consequently, a 95% confidence interval for µ is from
262.23
262.23
513.32 − 2.064 · √
to 513.32 + 2.064 · √
,
25
25
or 405.07 to 621.57. We can be 95% confident that the mean value lost, µ, of all
last year’s pick-pocket offenses is somewhere between $405.07 and $621.57.
EXAMPLE
8.10
Illustrates Procedure 8.2
The U.S. Department of Agriculture publishes data on chicken consumption in
Food Consumption, Prices, and Expenditures. Last year’s chicken consumptions,
in pounds, for 17 randomly selected people are displayed in Table 8.6. Use the data
to obtain a 90% confidence interval for last year’s mean chicken consumption, µ.
TABLE 8.6
Sample of last
year’s chicken
consumptions (lbs)
Solution
47
59
53
39
53
51
62
55
50
49
0
72
50
65
45
70
63
A normal probability plot of the data in Table 8.6 is displayed in Fig. 8.9(a). The
plot reveals an outlier—the observation of 0 lb. Since the sample size is only
moderate, it is inappropriate to apply Procedure 8.2 to the data in Table 8.6.