Two Samples: Tests on Two Proportions

10.9 Two Samples: Tests on Two Proportions

  Situations often arise where we wish to test the hypothesis that two proportions are equal. For example, we might want to show evidence that the proportion of doctors who are pediatricians in one state is equal to the proportion in another state. A person may decide to give up smoking only if he or she is convinced that the proportion of smokers with lung cancer exceeds the proportion of nonsmokers with lung cancer.

  In general, we wish to test the null hypothesis that two proportions, or bino-

  mial parameters, are equal. That is, we are testing p 1 =p 2 against one of the alternatives p 1

p 2 , or p 1 =p 2 . Of course, this is equivalent to testing the null hypothesis that p 1 −p 2 = 0 against one of the alternatives p 1 −p 2 < 0,

  p 1 −p 2 > 0, or p 1 −p 2 = 0. The statistic on which we base our decision is the

  random variable + P 1 −+ P 2 . Independent samples of sizes n 1 and n 2 are selected at random from two binomial populations and the proportions of successes + P 1 and + P 2

  for the two samples are computed.

  In our construction of confidence intervals for p 1 and p 2 we noted, for n 1 and n 2 sufficiently large, that the point estimator + P 1 minus + P 2 was approximately normally

  distributed with mean

  μ P 1 − P 2 =p 1 −p 2

  and variance

  Therefore, our critical region(s) can be established by using the standard normal variable

  When H 0 is true, we can substitute p 1 =p 2 = p and q 1 =q 2 = q (where p and

  q are the common values) in the preceding formula for Z to give the form

  pq(1n 1 + 1n 2 )

  To compute a value of Z, however, we must estimate the parameters p and q that appear in the radical. Upon pooling the data from both samples, the pooled estimate of the proportion p is

  x 1 +x 2

  p= ˆ

  Chapter 10 One- and Two-Sample Tests of Hypotheses

  where x 1 and x 2 are the numbers of successes in each of the two samples. Substi-

  tuting ˆ p for p and ˆ q=1 − ˆp for q, the z-value for testing p 1 =p 2 is determined

  from the formula

  pˆ ˆ q(1n 1 + 1n 2 )

  The critical regions for the appropriate alternative hypotheses are set up as before, using critical points of the standard normal curve. Hence, for the alternative

  p 1 =p 2 at the α-level of significance, the critical region is z < −z α2 or z > z α2 .

  For a test where the alternative is p 1

  when the alternative is p 1 >p 2 , the critical region is z > z α .

  Example 10.11:

  A vote is to be taken among the residents of a town and the surrounding county to determine whether a proposed chemical plant should be constructed. The con- struction site is within the town limits, and for this reason many voters in the county believe that the proposal will pass because of the large proportion of town voters who favor the construction. To determine if there is a significant difference in the proportions of town voters and county voters favoring the proposal, a poll is taken. If 120 of 200 town voters favor the proposal and 240 of 500 county residents favor it, would you agree that the proportion of town voters favoring the proposal is higher than the proportion of county voters? Use an α = 0.05 level of significance.

  Solution : Let p 1 and p 2 be the true proportions of voters in the town and county, respectively,

  favoring the proposal.

  4. Critical region: z > 1.645.

  = 0.60, p ˆ 2 =

  n 1 +n 2 200 + 500 Therefore,

  (0.51)(0.49)(1200 + 1500) P = P (Z > 2.9) = 0.0019.

  6. Decision: Reject H 0 and agree that the proportion of town voters favoring

  the proposal is higher than the proportion of county voters.

  Exercises

  Exercises

  10.55 A marketing expert for a pasta-making com- 10.62 In a controlled laboratory experiment, scien-

  pany believes that 40 of pasta lovers prefer lasagna. tists at the University of Minnesota discovered that If 9 out of 20 pasta lovers choose lasagna over other pas-

  25 of a certain strain of rats subjected to a 20 coffee

  tas, what can be concluded about the expert’s claim? bean diet and then force-fed a powerful cancer-causing Use a 0.05 level of significance.

  chemical later developed cancerous tumors. Would we have reason to believe that the proportion of rats devel-

  10.56 Suppose that, in the past, 40 of all adults oping tumors when subjected to this diet has increased favored capital punishment. Do we have reason to if the experiment were repeated and 16 of 48 rats de- believe that the proportion of adults favoring capital veloped tumors? Use a 0.05 level of significance. punishment has increased if, in a random sample of 15

  adults, 8 favor capital punishment? Use a 0.05 level of 10.63 In a study to estimate the proportion of resi-

  significance.

  dents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that

  63 of 100 urban residents favor the construction while

  10.57 A new radar device is being considered for a only 59 of 125 suburban residents are in favor. Is there certain missile defense system. The system is checked

  a significant difference between the proportions of ur-

  by experimenting with aircraft in which a kill or a no ban and suburban residents who favor construction of kill is simulated. If, in 300 trials, 250 kills occur, accept the nuclear plant? Make use of a P -value. or reject, at the 0.04 level of significance, the claim that

  the probability of a kill with the new system does not 10.64 In a study on the fertility of married women

  exceed the 0.8 probability of the existing device.

  conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups of childless

  10.58 It is believed that at least 60 of the residents wives aged 25 to 29 were selected at random, and each in a certain area favor an annexation suit by a neigh- was asked if she eventually planned to have a child. boring city. What conclusion would you draw if only One group was selected from among wives married 110 in a sample of 200 voters favored the suit? Use a less than two years and the other from among wives

  0.05 level of significance.

  married five years. Suppose that 240 of the 300 wives married less than two years planned to have children

  10.59 A fuel oil company claims that one-fifth of the some day compared to 288 of the 400 wives married homes in a certain city are heated by oil. Do we have five years. Can we conclude that the proportion of reason to believe that fewer than one-fifth are heated wives married less than two years who planned to have by oil if, in a random sample of 1000 homes in this city, children is significantly higher than the proportion of 136 are heated by oil? Use a P -value in your conclu- wives married five years? Make use of a P -value. sion.

  10.65 An urban community would like to show that

  10.60 At a certain college, it is estimated that at most the incidence of breast cancer is higher in their area 25 of the students ride bicycles to class. Does this than in a nearby rural area. (PCB levels were found to seem to be a valid estimate if, in a random sample of

  be higher in the soil of the urban community.) If it is

  90 college students, 28 are found to ride bicycles to found that 20 of 200 adult women in the urban com- class? Use a 0.05 level of significance.

  munity have breast cancer and 10 of 150 adult women in the rural community have breast cancer, can we con- clude at the 0.05 level of significance that breast cancer

  10.61 In a winter of an epidemic flu, the parents of is more prevalent in the urban community? 2000 babies were surveyed by researchers at a well- known pharmaceutical company to determine if the

  10.66 Group Project : The class should be divided

  company’s new medicine was effective after two days. into pairs of students for this project. Suppose it is Among 120 babies who had the flu and were given the conjectured that at least 25 of students at your uni- medicine, 29 were cured within two days. Among 280 versity exercise for more than two hours a week. Col- babies who had the flu but were not given the medicine, lect data from a random sample of 50 students. Ask

  56 recovered within two days. Is there any significant each student if he or she works out for at least two indication that supports the company’s claim of the hours per week. Then do the computations that allow effectiveness of the medicine?

  either rejection or nonrejection of the above conjecture. Show all work and quote a P -value in your conclusion.

  Chapter 10 One- and Two-Sample Tests of Hypotheses