OVERVIEW: This section, although notlengthy in terms of pages, has lots of "meat" in it. There are twostandard error formulas that are commonly used when comparingproportions from two independent samples. Both formula are listed onthe formula sheet that is provided to those who take the AdvancedPlacement Statistics Examination. On page 679, the authors remindreaders that "variances add" when you are talking about differences.This thought is important in understanding the standard error formulalisted on page 681. Just a reminder... the "variances add" idea goesback to page 400. Review, if necessary.
I'll attempt to summarize this important sectionwith two examples:
Example 1 (In whichwe assume that there is no reason to assume that the samples comefrom populations with equal variances.) We will use formulas shown onpage 681.
Do voters want State Proposition A?
# wanting Prop. A
N1 = 93
p1(hat) = .796
N2 = 87
p2(hat) = .621
SE = sqrt[(.796)(1-.796)/93 + (.621)(1-.621)/87] =0.06672
Since the sampling distribution ofp1(hat) -p2(hat)isapproximately normal, the 95% confidence interval forp1 -p2 is
(.796-.621) plus/minus 1.96(0.06672) = 0.174plus/minus 0.1308 = [4.32%, 30.48%].
Since this is a 95% confidence interval, we knowthat 19 out of 20 confidence intervals constructed in this fashionwould contain the true percentage difference. In other words, it isstatistically reasonable to assume that the actual value for theparameter p1- p2 isbetween 4% and 30%. Since this interval does not contain 0%, it isreasonable to conclude that these samples came from populations thatdiffer in their support for Proposition A.
Example 2 (In whichwe initially assume that the samples come from populations with equalvariances.) We will use formulas shown on page 684.
Assume we have the following statistical resultsfrom a research project.
N1 = 1,865
p1(hat) = .0552
N2 = 1,712
p2(hat) = .0461
Null hypothesis H
0: p 1= p 2
Alternate hypothesis Ha: p
1is not equal top2.
Pooling the data, we obtain p(hat) =(103+79)/(1895+1712) = .0509.
z = (.0552-.0461)
/sqrt[(.0509)(1-.0509)(1/1865 +1/1712)] = 1.24.
At the 5% level of significance (2-tail), thecritical region is z < -1.96 or z > 1.96. Since our calculatedz of 1.24 is not in this region, we would not rejectH0.
Alternate approach:Using the TI-83, normalcdf(
1.24,1E99,0,1) = .1074877610. Since this is a 2-tailsituation, the P-value is 2(.1074877610), or about 21.5%. There isnot strong evidence to reject H 0.
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