o_18ufhu7n51jij1pedr1jjeu19lpa.pdf

350 Chapter 10 Statistical Inferences about Two Populations
Rejection Region
z .001 = –3.08
Observed z = –10.42
x 1 – x 2 = –701.9
x 1 – x 2 = –2,375
z = 0.0
μ1 – μ 2 = 0.0
z
x 1 – x 2
The probability of obtaining an observed z value of -10.42 by chance is virtually
zero, because the value is beyond the limits of the z table. By the p-value, the null
hypothesis is rejected because the probability is .0000, or less than a = .001.
If this problem were worked by the critical value method, what critical value of
the difference in the two means would have to be surpassed to reject the null hypothesis
for a table z value of -3.08? The answer is
(x 1 - x 2 ) c = (m 1 - m 2 ) - z s 2 1
+ s2 2
A n 1 n 2
= 0 - 3.08(227.9) = -701.9
The difference in sample means would need to be at least 701.9 to reject the null
hypothesis. The actual sample difference in this problem was -2375 (3352 - 5727),
which is considerably larger than the critical value of difference. Thus, with the critical
value method also, the null hypothesis is rejected.
Confidence Intervals
Sometimes being able to estimate the difference in the means of two populations is valuable.
By how much do two populations differ in size or weight or age? By how much do two
products differ in effectiveness? Do two different manufacturing or training methods produce
different mean results? The answers to these questions are often difficult to obtain
through census techniques. The alternative is to take a random sample from each of the two
populations and study the difference in the sample means.
Algebraically, formula 10.1 can be manipulated to produce a formula for constructing
confidence intervals for the difference in two population means.
CONFIDENCE INTERVAL TO
s 2 1
ESTIMATE m 1 - m 2 (10.2)
(x 1 - x 2 ) - z
A n
+ s2 2
s 2 1
… m
1 n 1 - m 2 … (x 1 - x 2 ) + z
2 A n
+ s2 2
1
n 2
Suppose a study is conducted to estimate the difference between middle-income shoppers
and low-income shoppers in terms of the average amount saved on grocery bills per
week by using coupons. Random samples of 60 middle-income shoppers and 80 lowincome
shoppers are taken, and their purchases are monitored for one week. The average
amounts saved with coupons, as well as sample sizes and population standard deviations
are in the table on the next page.