Inferences About Two Population Proportions

Transcription

1 Inferences About Two Population Proportions MATH 130, Elements of Statistics I J. Robert Buchanan Department of Mathematics Fall 2018

2 Background Recall: for a single population the sampling proportion ˆp is normally distributed with mean µˆp = p and standard deviation p(1 p) σˆp = provided that n p(1 p) 10. n

3 Background Recall: for a single population the sampling proportion ˆp is normally distributed with mean µˆp = p and standard deviation p(1 p) σˆp = provided that n p(1 p) 10. n Test statistic: for a hypothesis test about a single population proportion, z = ˆp p. p(1 p) n

4 Two Population Proportions Suppose a simple random sample of size n 1 is taken from a population where x 1 individuals have a certain characteristic and suppose a simple random sample of size n 2 is taken from a different population where x 2 individuals have the same characteristic, then the sampling distribution of ˆp 1 ˆp 2 is approximately normal with mean and standard deviation σˆp1 ˆp 2 = µˆp1 ˆp 2 = p 1 p 2 p 1 (1 p 1 ) n 1 + p 2(1 p 2 ) n 2 provided n 1ˆp 1 (1 ˆp 1 ) 10 and n 2ˆp 2 (1 ˆp 2 ) 10.

5 Normalized ˆp 1 ˆp 2 Since then µˆp1 ˆp 2 = p 1 p 2 σˆp1 ˆp 2 = p 1 (1 p 1 ) + p 2(1 p 2 ) n 1 n 2 z = (ˆp 1 ˆp 2 ) (p 1 p 2 ) p1 (1 p 1 ) n 1 + p 2(1 p 2 ) n 2 is normally distributed with mean 0 and standard deviation 1.

6 Hypothesis Testing During hypothesis tests on two population proportions which implies H 0 : p 1 = p 2 z = (ˆp 1 ˆp 2 ) (p 1 p 2 ) p1 (1 p 1 ) n 1 + p 2(1 p 2 ) n 2 (ˆp 1 ˆp 2 ) = p1 (1 p 1 ) n 1 + p 2(1 p 2 ) n 2

7 Pooled Estimate (1 of 2) If p 1 = p 2 then we do not need the subscripts and we may say p 1 = p 2 = p. Thus z = = = (ˆp 1 ˆp 2 ) p1 (1 p 1 ) n 1 + p 2(1 p 2 ) n 2 (ˆp 1 ˆp 2 ) p(1 p) n 1 + p(1 p) n 2 (ˆp 1 ˆp 2 ) p(1 p) 1 n n 2

8 Pooled Estimate (2 of 2) However p (the common population proportion) is unknown, so we will use a point estimate of p called the pooled estimate of p: ˆp = x 1 + x 2 n 1 + n 2. So finally, z = (ˆp 1 ˆp 2 ) ˆp(1 ˆp) 1 n n 2.

9 Steps in Hypothesis Testing 1. Determine the null and alternative hypotheses. Two-tailed Left-tailed Right-tailed H 0 : p 1 = p 2 H 0 : p 1 = p 2 H 0 : p 1 = p 2 H 1 : p 1 p 2 H 1 : p 1 < p 2 H 1 : p 1 > p 2 2. Select a level of significance α. 3. Compute the test statistic: z 0 = (ˆp 1 ˆp 2 ) ˆp(1 ˆp) 1 n n 2 4. Use the classical or P-value approach to make a decision. 5. State the conclusion.

10 Assumptions We have made the following assumptions in the process of performing a hypothesis test on two population proportions: The samples are independently obtained using simple random sampling. n 1ˆp 1 (1 ˆp 1 ) 10 and n 2ˆp 2 (1 ˆp 2 ) 10. n N 1 and n N 2.

11 Example (Classical Approach, 1 of 3) A salesman for a new manufacturer of cell phones claims not only that they cost less but also that the percentage of defective phones is no higher than those of a competitor. To test this claim a retailer took random samples of each of the two product lines. Test the salesman s claim at the 0.05 level of significance. Product Number Defective Number Checked Salesman s Competitor s 6 150

12 Example (Classical Approach, 2 of 3) H 0 : p 1 = p 2 H 1 : p 1 < p 2 (left-tailed test) α = 0.05, z α = ˆp 1 = 0.10 ˆp 2 = 0.04 ˆp = = 0.07 Test statistic: z 0 = (ˆp 1 ˆp 2 ) ˆp(1 ˆp) 1 n n 2 ( ) = 0.07(1 0.07) = 2.04

13 Example (Classical Approach, 3 of 3) z 0 =2.04 z α = Decision: do not reject H 0. Conclusion: the sample data do not support the salesman s claim that the defect rate of his cell phones is no higher than the defect rate of a competitor s phones.

14 Example (P-Value Approach, 1 of 3) Suppose a study is conducted to compare the service provided by manufacturers to home PC owners and work PC owners. Of 220 home PC owners who had trouble, 98 reported that their problem was not resolved satisfactorily. Of 180 work PC owners who experienced difficulty, 52 reported that the problem was not resolved. At the 0.05 level of significance, did the home PC owners experience more problems that could not be resolved by the manufacturer?

15 Example (P-Value Approach, 2 of 3) H 0 : p 1 = p 2 H 1 : p 1 > p 2 (right-tailed test) α = 0.05, z α = ˆp 1 = ˆp 2 = ˆp = = Test statistic: z 0 = (ˆp 1 ˆp 2 ) ˆp(1 ˆp) 1 n n 2 ( ) = 0.375( ) = 3.21

16 Example (P-Value Approach, 3 of 3) Decision: reject H 0. P-value = P(z > 3.21) = 1 P(z < 3.21) = = < 0.05 = α Conclusion: the sample data support that home PC owners experience more problems that could not be resolved by the manufacturer than work PC owners.

17 Confidence Interval Estimates Assuming again that the samples are independently obtained using simple random sampling, n 1ˆp 1 (1 ˆp 1 ) 10 and n 2ˆp 2 (1 ˆp 2 ) 10, and n N 1 and n N 2. The (1 α) 100% confidence interval estimate of p 1 p 2 is ˆp 1 (1 ˆp 1 ) Lower bound: (ˆp 1 ˆp 2 ) z α/2 + ˆp 2(1 ˆp 2 ) n 1 n 2 ˆp 1 (1 ˆp 1 ) Upper bound: (ˆp 1 ˆp 2 ) + z α/2 + ˆp 2(1 ˆp 2 ) n 1 n 2

18 Example (1 of 3) Smoking poses an added health risk among people with diabetes. The journal Science News investigated the smoking rates for male and female diabetics. Gender n Number who Smoke Male Female Construct the 98% confidence interval estimate of the difference in the smoking rates between male and female diabetics.

19 Example (2 of 3) ˆp 1 = 215/500 = ˆp 2 = 170/500 = Margin of Error: α = 0.02 α/2 = 0.01 z α/2 = ˆp 1 (1 ˆp 1 ) E = z α/2 + ˆp 2(1 ˆp 2 ) n 1 n ( ) = = ( ) 500

20 Example (3 of 3) 98% Confidence Interval: ([ˆp 1 ˆp 2 ] E, [ˆp 1 ˆp 2 ] + E) = ([ ] 0.071, [ ] ) = (0.019, 0.161)

21 Estimating Sample Size Suppose we want to collect data to construct a confidence interval.

22 Estimating Sample Size Suppose we want to collect data to construct a confidence interval. Question: How large a sample from each population should we gather?

23 Estimating Sample Size Suppose we want to collect data to construct a confidence interval. Question: How large a sample from each population should we gather? We will assume that n 1 = n 2 = n, then ˆp 1 (1 ˆp 1 ) E = z α/2 + ˆp 2(1 ˆp 2 ) n 1 n 2 = z α/2 ˆp1 (1 ˆp 1 ) + ˆp 2 (1 ˆp 2 ) n

24 Sample Size If prior estimates of ˆp 1 and ˆp 2 are available n = [ˆp 1 (1 ˆp 1 ) + ˆp 2 (1 ˆp 2 )] ( zα/2 E ) 2. If no prior estimates are available n = 0.5 ( zα/2 E ) 2.

25 Example (1 of 3) You wish to conduct a survey of male vs. female voters preferences for a political candidate. Since preferences change from day to day such a survey must be conducted repeatedly in the days leading up to an election. Suppose that yesterday 45% of male voters preferred a candidate and 55% of female voters preferred the same candidate. How large a sample should be collected to estimate the difference in the male vs. female voter preference with a margin of error of 3% at the 95% confidence level?

26 Example (2 of 3) We are told prior estimates of ˆp 1 and ˆp 2 : Therefore ˆp 1 = 0.45 ˆp 2 = 0.55 E = 0.03 α = 0.05 α/2 = z α/2 = n = [ˆp 1 (1 ˆp 1 ) + ˆp 2 (1 ˆp 2 )] ( ) 2 zα/2 = [0.45(1 0.45) (1 0.55)] = E ( )

27 Example (3 of 3) If no prior estimate of voter preference were available then the sample size would be ( ) 2 zα/2 n = 0.5 E ( ) = =