1. 9.2a Tests about a Population Proportion Target Goal: I can check the conditions for carrying out a test about a population proportion. I can perform a significance test for a sample proportion. h.w: pg 548: 27 – 30, pg 562: 41, 43, 45

2.  Carrying Out a Significance Test Tests About a Population Proportion Suppose a basketball player who claimed to be an 80% free-throw shooter. In an SRS of 50 free-throws, he made 32. His sample proportion of made shots, 32/50 = 0.64, is much lower than what he claimed. Does it provide convincing evidence against his claim? To find out, we must perform a significance test of H 0 : p = 0.80 H a : p < 0.80 where p = the actual proportion of free throws the shooter makes in the long run.

3.  For a hypothesis test where H o : p = p o, use p o to estimate p.  For a confidence interval: we used as an estimate of. we used as an estimate of.

4. Assumptions for Inference about a Proportion: 1. Random: SRS 2. Independent: Population  10n 3. Normal:  10 and  10 for a hypothesis test.  10 and  10 for a hypothesis test. for a confidence interval. for a confidence interval.

5.  To test hypothesis H o : p = p o,

6.  The One-Sample z Test for a Proportion The z statistic has approximately the standard Normal distribution when H 0 is true. P -values therefore come from the standard Normal distribution. Here is a summary of the details for a one-sample z test for a proportion. Tests About a Population Proportion Choose an SRS of size n from a large population that contains an unknown proportion p of successes. To test the hypothesis H 0 : p = p 0, compute the z statistic Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis H a : Choose an SRS of size n from a large population that contains an unknown proportion p of successes. To test the hypothesis H 0 : p = p 0, compute the z statistic Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis H a : One-Sample z Test for a Proportion Use this test only when the expected numbers of successes and failures np 0 and n(1 - p 0 ) are both at least 10 and the population is at least 10 times as large as the sample. Use this test only when the expected numbers of successes and failures np 0 and n(1 - p 0 ) are both at least 10 and the population is at least 10 times as large as the sample. 0

7. Ex. Binge Drinking in College  In a representative of 140 colleges and 17592 students, 7741 students identify themselves as binge drinkers. Considering this SRS, does this constitute strong evidence that more than 40% of all college students engage in binge drinking?

8. Step 1: State - What hypotheses do you want to test, and at what significance level? State the hypothesis in words and symbols.  We want to test a claim about the proportion of all U.S. college students who have engaged in binge drinking at the α =.05 level. Our hypotheses are  H o : p =.40 40% of all college students are binge drinkers  H a : p >.40 more than 40% of all U.S. college students have engaged in binge drinking.

9. Step 2: Plan - Choose the appropriate inference procedure. Verify the conditions for using the selected procedure. To test claim, we will use the one prop z test. Check conditions.  Random: SRS? We are given SRS.  Independent: Total population > 10 n: 10(17592) = there are more than 175,920 college students in the country. ( to use sample σ) yes independent.  Normal: np o and nq o ≥ 10? 17592(.40) = 7036.8 ≥ 10 17592(.60) = 10,555.2 ≥ 10 Yes, we can use normal approximation.

10. Step 3: Do - If the conditions are met, carry out the inference procedure.  Calculate z statistic  With a z score this large, the P-value is approx. 0.

11. Step 4. Conclude - Interpret the results in the context of the problem  The p-value (= 0) this small, < α =.05, tells us that we have no chance of obtaining a sample proportion  We reject H o and conclude that more than 40% of U.S. college students have engaged in binge drinking.

12. Ex. Is that Coin Fair?  The French naturalist Count Buffon tossed a coin 4040 times and counted 2048 heads. The sample proportion of heads is  = 0.5069  Is this evidence that Buffon’s coin was not balanced?

13. Step 1: State  The parameter p is the probability of tossing a head. The population contains the results of tossing the coin forever.  Our hypotheses are:  H o : p =.50 The coin is balanced.  H a : p 0.5 Buffon’s coin is not balanced. The null hypothesis gives p the value p o =.50.

14. Step 2: Plan We will use the one prop z test. Check conditions.  SRS? Yes  Total population > 10 n?  np o and nq o ≥ 10? np o = 4040(0.5) = 2020 ≥ 10 nq o = 4040(0.5) = 2020 ≥ 10 The population of tosses is infinite.

15. Step 3: Do - If the conditions are met, carry out the inference procedure.  Calculate z statistic  P-value = (two sided so)  Normcdf(.88,E99) 2(.1894) =.3788

16. Step 4. Conclude - Interpret the results in the context of the problem.  The proportion of heads as far away from 1/2 as Buffon’s would happen 38% of the time. This provides little evidence against H o. Thus, Buffon’s result doesn’t show that his coin is unbalanced.

17.  Note: calculator much quicker.  STAT: TESTS:1-Prop Z Test  Read pg. 549 - 555