1. Slide Slide 1 Section 8-3 Testing a Claim About a Proportion

2. Slide Slide 2 Key Concept This section presents complete procedures for testing a hypothesis (or claim) made about a population proportion. This section uses the components introduced in the previous section for the P-value method, the traditional method or the use of confidence intervals.

3. Slide Slide 3 1) The sample observations are a simple random sample. 2) The conditions for a binomial distribution are satisfied (Section 5-3). 3) The conditions np  5 and nq  5 are satisfied, so the binomial distribution of sample proportions can be approximated by a normal distribution with µ = np and  = npq. Requirements for Testing Claims About a Population Proportion p

4. Slide Slide 4 Notation p = population proportion (used in the null hypothesis) q = 1 – p  n = number of trials p = x (sample proportion) n

5. Slide Slide 5 p – pp – p pq n z =z =  Test Statistic for Testing a Claim About a Proportion

6. Slide Slide 6 P-Value Method Use the same method as described in Section 8-2 and in Figure 8-8. Use the standard normal distribution (Table A-2).

7. Slide Slide 7 Traditional Method Use the same method as described in Section 8-2 and in Figure 8-9.

8. Slide Slide 8 Confidence Interval Method Use the same method as described in Section 8-2 and in Table 8-2.

9. Slide Slide 9 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56.  np = (880)(0.5) = 440  5 nq = (880)(0.5) = 440  5

10. Slide Slide 10 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P-value Method.  Referring to Table A-2, we see that for values of z = 3.50 and higher, we use 0.9999 for the cumulative area to the left of the test statistic. The P-value is 1 – 0.9999 = 0.0001. H 0 : p = 0.5 H 1 : p > 0.5  = 0.05 pq n p – p z =z =  0.56 – 0.5 (0.5)(0.5) 880 = = 3.56

11. Slide Slide 11 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P-value Method.  Since the P-value of 0.0001 is less than the significance level of  = 0.05, we reject the null hypothesis. There is sufficient evidence to support the claim. H 0 : p = 0.5 H 1 : p > 0.5  = 0.05 pq n p – p z =z =  0.56 – 0.5 (0.5)(0.5) 880 = = 3.56

12. Slide Slide 12 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the P-value Method.  H 0 : p = 0.5 H 1 : p > 0.5  = 0.05 z = 3.56

13. Slide Slide 13 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the Traditional Method.  H 0 : p = 0.5 H 1 : p > 0.5  = 0.05 pq n p – p z =z =  0.56 – 0.5 (0.5)(0.5) 880 = = 3.56 This is a right-tailed test, so the critical region is an area of 0.05. We find that z = 1.645 is the critical value of the critical region. We reject the null hypothesis. There is sufficient evidence to support the claim.

14. Slide Slide 14 Example: An article distributed by the Associated Press included these results from a nationwide survey: Of 880 randomly selected drivers, 56% admitted that they run red lights. The claim is that the majority of all Americans run red lights. That is, p > 0.5. The sample data are n = 880, and p = 0.56. We will use the confidence interval method.  For a one-tailed hypothesis test with significance level , we will construct a confidence interval with a confidence level of 1 – 2 . We construct a 90% confidence interval. We obtain 0.533 0.5.

15. Slide Slide 15 CAUTION When testing claims about a population proportion, the traditional method and the P-value method are equivalent and will yield the same result since they use the same standard deviation based on the claimed proportion p. However, the confidence interval uses an estimated standard deviation based upon the sample proportion p. Consequently, it is possible that the traditional and P-value methods may yield a different conclusion than the confidence interval method. A good strategy is to use a confidence interval to estimate a population proportion, but use the P-value or traditional method for testing a hypothesis. 

16. Slide Slide 16 (determining the sample proportion of households with cable TV) p = = = 0.64 x n 96 (96+54)  and 54 do not” is calculated using p sometimes must be calculated “96 surveyed households have cable TV   p sometimes is given directly “10% of the observed sports cars are red” is expressed as p = 0.10  Obtaining P 

17. Slide Slide 17 Example: When Gregory Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods. According to Mendel’s theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level with the P-value method to test the claim that the proportion of peas with yellow pods is equal to 1/4. We note that n = 428 + 152 = 580, so p = 0.262, and p = 0.25. 

18. Slide Slide 18 Example: When Gregory Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods. According to Mendel’s theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level with the P-value method to test the claim that the proportion of peas with yellow pods is equal to 1/4. H 0 : p = 0.25 H 1 : p  0.25 n = 580  = 0.05 p = 0.262  0.262 – 0.25 (0.25)(0.75) 580 = = 0.67 z =z = p – p pq n  Since this is a two-tailed test, the P-value is twice the area to the right of the test statistic. Using Table A-2, z = 0.67 is 1 – 0.7486 = 0.2514.

19. Slide Slide 19 Example: When Gregory Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 peas with yellow pods. According to Mendel’s theory, 1/4 of the offspring peas should have yellow pods. Use a 0.05 significance level with the P-value method to test the claim that the proportion of peas with yellow pods is equal to 1/4. The P-value is 2(0.2514) = 0.5028. We fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that 1/4 of the peas have yellow pods. H 0 : p = 0.25 H 1 : p  0.25 n = 580  = 0.05 p = 0.262 0.262 – 0.25 (0.25)(0.75) 580 = = 0.67 z =z = p – p pq n  

20. Slide Slide 20 Recap In this section we have discussed:  Test statistics for claims about a proportion.  P-value method.  Confidence interval method.  Obtaining p.