Section 9.3 Testing a Proportion p

2 Focus Points Identify the components needed for testing a proportion. Compute the sample test statistic. Find the P-value and conclude the test.

3 Testing a Proportion p Throughout this section, we will assume that the situations we are dealing with satisfy the conditions underlying the binomial distribution. In particular, we will let r be a binomial random variable. This means that r is the number of successes out of n independent binomial trials. We will use = r/n as our estimate for p, the population probability of success on each trial.

4 Testing a Proportion p The letter q again represents the population probability of failure on each trial, and so q = 1 – p. We also assume that the samples are large (i.e., np > 5 and nq > 5). For large samples, np > 5, and nq > 5, the distribution of = r/n values is well approximated by a normal curve with mean  and standard deviation  as follows:  = p

5 Testing a Proportion p The null and alternate hypotheses for tests of proportions are Left-Tailed Test Right-Tailed Test Two-Tailed Test H 0 : p = k H o : p = k H 0 : p = k H 1 : p k H 1 : p ≠ k depending on what is asked for in the problem. Notice that since p is a probability, the value k must be between 0 and 1.

6 Testing a Proportion p For tests of proportions, we need to convert the sample test statistic to a z value. Then we can find a P-value appropriate for the test. The distribution is approximately normal, with mean p and standard deviation. Therefore, the conversion of to z follows the formula

7 Testing a Proportion p Using this mathematical information about the sampling distribution for, the basic procedure is similar to tests you have conducted before.

8 Testing a Proportion p Procedure:

9 Testing a Proportion p cont’d

10 Example 6 – Testing p A team of eye surgeons has developed a new technique for a risky eye operation to restore the sight of people blinded from a certain disease. Under the old method, it is known that only 30% of the patients who undergo this operation recover their eyesight. Suppose that surgeons in various hospitals have performed a total of 225 operations using the new method and that 88 have been successful (i.e., the patients fully recovered their sight). Can we justify the claim that the new method is better than the old one? (Use a 1% level of significance.)

11 Example 6 – Solution (a) Establish H 0 and H 1 and note the level of significance. The level of significance is  = Let p be the probability that a patient fully recovers his or her eyesight. The null hypothesis is that p is still 0.30, even for the new method. The alternate hypothesis is that the new method has improved the chances of a patient recovering his or her eyesight. Therefore, H 0 : p = 0.30 and H 1 : p > 0.30

12 Example 6 – Solution (b) Check Requirements Is the sample sufficiently large to justify use of the normal distribution for ? Find the sample test statistic and convert it to a z value, if appropriate. Using p from H 0 we note that np = 225(0.3) = 67.5 is greater than 5 and that nq = 225(0.7) = is also greater than 5, so we can use the normal distribution for the sample statistic. cont’d

13 Example 6 – Solution The z value corresponding to is In the formula, the value for p is from the null hypothesis. H 0 specifies that p = 0.30, so q = 1 – 0.30 = cont’d

14 Example 6 – Solution (c) Find the P-value of the test statistic. Figure 8-10 shows the P-value. Since we have a right tailed test, the P-value is the area to the right of z = 2.95 Figure 8.10 P-value Area cont’d

15 Example 6 – Solution Using the normal distribution (Table 5 of Appendix II), we find that P-value = P(z > 2.95)  Calculator: normalcdf(2.95, 10, 0, 1) = (d) Conclude the test. Since the P-value of  0.01 for , we reject H 0. (e) Interpretation Interpret the results in the context of the problem. At the 1% level of significance, the evidence shows that the population probability of success for the new surgery technique is higher than that of the old technique. cont’d

16 Calculator STAT  TESTS  5: 1-PropTest P0: 0.30 x: 88 n: 225 Prop≠P0 p0 Calculate Draw z= p=

17 Guided Exercise 5: Testing p A botanist has produced a new variety of hybrid wheat that is better able to withstand drought than other varieties. The botanist knows that for the parent plants, the proportion of seeds germinating is 80%. The proportion of seeds germinating for the hybrid variety is unknown, but the botanist claims that it is 80%. To test this claim, 400 seeds from the hybrid plant are tested, and it is found that 312 germinate. Use a 5% level of significance to test the claim that the proportion germinating for the hybrid is 80%.

18 Guided Exercise 5: Testing p (a) Let p be the proportion of hybrid seeds that will germinate. Notice that we have no prior knowledge about the germination proportion for the hybrid plant. State H 0 and H 1. What is the required level of significance? H 0: μ = 0.80 H 1 : μ ≠ 0.80; α = 0.05 (b) Calculate the sample test statistic. Using the value of p in H 0, are both np>75 and nq>75? Can we use the normal distribution for ? 312/400 = 0.78

19 Guided Exercise 5: Testing p c) Next, we convert the sample test statistic = 0.78 to a z value. Based on our choice for H0, what value should we use for p in our formula? Since q = 1 - p, what value should we use for q? Using these values for p and q, convert to a z value. Z = (d) Is the test right-tailed, left-tailed, or two-tailed? Find the P-value of the sample test statistic and sketch a standard normal curve showing the P-value. 2  normalcdf(-10, -1.00, 0, 1) =

20 Guided Exercise 5: Testing p e) Do we reject or fail to reject H 0 ? Interpret your conclusion in the context of the application. Since P-value of > 0.05 for α we fail to reject H0. At the 5% level of significance, there is insufficient evidence to conclude that the botanist is wrong.

21 Calculator STAT  TESTS  5: 1-PropTest P0: 0.80 x: 312 n: 400 Prop: ≠P0 p0 Calculate Draw z=-1 p=

22 Answers to some even hw: 6. P-value ≈ P-value ≈ P-value ≈ P-value ≈ P-value ≈