WARM - UP Is the coin Fair? With p = The true proportion of times the coin lands on “Heads”, construct and interpret the 95% Confidence Interval for the following sample statistic. You flip the coin 220 times and find 59.5% of the time it lands heads. One Proportion z – Conf. Interval Random – The data was collected randomly by a chance event Appr. Norm: 220(0.595)=130.9 ≥ 10 AND 220(1–(0.595)) ≥ 10 Population of potential tosses ≥ 10 · 220 = 2200 We can be 95% confident that the true proportion of coin tosses landing on head is between 0.531 and 0.660.

TEXTBOOK PAGE 469: 1 – 7 1. Hypotheses. a) H0 : The governor’s “negatives” are 30%. (p = 0.30) HA : The governor’s “negatives” are less than 30%. (p < 0.30) b) H0 : The proportion of heads is 50%. (p = 0.50) HA : The proportion of heads is not 50%. (p ≠ 0.50) c) H0 : The proportion of people who quit smoking is 20%. (p = 0.20) HA : The proportion of people who quit smoking is greater than 20%. (p > 0.20) 2. More hypotheses. a) H0 : The proportion of high school graduates is 40%. (p = 0.40) HA : The proportion of high school graduates is not 40%. (p ≠ 0.40) b) H0 : The proportion of cars needing transmission repair is 20%. (p = 0.20) HA : The proportion of cars needing transmission repair is less than 20%. (p < 0.20) c) H0 : The proportion of people who like the flavor is 60%. (p = 0.60) HA : The proportion of people who like the flavor is greater than 60%. (p > 0.60)

d 5. Relief. It is not reasonable to conclude that the new formula and the old one are equally effective. Furthermore, our inability to make that conclusion has nothing to do with the P-value. We can not prove the null hypothesis (that the new formula and the old formula are equally effective), but can only fail to find evidence that would cause us to reject it. All we can say about this P-value is that there is a 27% chance of seeing the observed effectiveness from natural sampling variation if the new formula and the old one are equally effective.

6. Cars. It is reasonable to conclude that a greater proportion of high schoolers have cars. If the proportion were no higher than it was a decade ago, there is only a 1.7% chance of seeing such a high sample proportion just from natural sampling variability. 7. He cheats! a) Two losses in a row aren’t convincing. There is a 25% chance of losing twice in a row, and that is not unusual. b) If the process is fair, three losses in a row can be expected to happen about 12.5% of the time. (0.5)(0.5)(0.5) = 0.125. c) Three losses in a row is still not a convincing occurrence. We’d expect that to happen about once every eight times we tossed a coin three times. d) Answers may vary. Maybe 5 times would be convincing. The chances of 5 losses in a row are only 1 in 32, which seems unusual.

Ch. 20: Hypothesis Tests (Proportions) PARAMETER: p = True proportion of… H0: p = p0 Ha: p < , >, or ≠ p0 One Proportion z-Test OR Assumptions (prop.): 1. SRS – Must be stated 2. Appr. Norm: np ≥ 10 n(1 – p) ≥ 10 3. Population ≥ 10n z-Test Stat. → P – Value Ha: p < p0 Normalcdf (-E99,z) Ha: p > p0 Normalcdf (z, E99) Ha: p ≠ p0 2 x Normalcdf (|z|, E99) 6. Reject or Fail to Reject H0 (based on .05) Conclusion in Context:

Statistics Hypothesis Test PHANTOMS P Parameter H Hypotheses A Assumptions N Name the test T Test statistic O Obtain p-value M Make decision S State conclusion in context

p = The true proportion of college students who binge drink. EXAMPLE: In 1995 Harvard studied the use of alcohol among college students. Binge Drinking (consuming more than 5 drinks on one occasion) was a concern. A SRS of 17592 students from the nation was taken with 7741 identifying themselves as binge drinkers. Does this constitute strong evidence that more than 40% of students binge drink? One Proportion z – Test p = The true proportion of college students who binge drink. H0: p = 0.40 Ha: p > 0.40 SRS – Stated Appr. Norm: 17592(.4)=7036.8 ≥10 17592(1-.4)=10555.2≥ 10 Since the P-Value is less than 0.05, there is strong evidence to REJECT H0 . More than 40% of college students binge drink.

One Proportion z – Test p = The true proportion of college students who have mothers who grad. college. H0: p = 0.31 Ha: p ≠ 0.31 Random – Assume Randomness Appr. Norm: 8368(.31)=2594… ≥10 8368(1-.31)=5774…≥ 10 Since the P-Value is less than 0.05, `REJECT H0 . There is evidence of a change in education level’s among mother.

Is the coin Fair? You suspect that you have a weighted coin. You flip the coin 220 times and find 124 heads. Is there enough statistical evidence to declare the coin unfair? Let p = the true proportion of coin tosses landing on heads. Ho: p = 0.5 Ha: p ≠ 0.5 One Proportion z – Test SRS – The data was collected randomly by a chance event Appr. Normal: 220 (0.5)=110 ≥ 10 AND 220 (1 – (0.5))=110 ≥ 10 Population of potential tosses ≥ 10 · 220 = 2200 Since our p-value is greater than 0.05 we “FAIL to REJECT H0“. There is not sufficient evidence to conclude the coin is unfair.

WHAT PERCENT OF THE EARTH IS COVERED WITH WATER?

One Proportion Significance Test Is there evidence that the proportion of the earth that is covered with water is different than 0.73 p = The true proportion of earth’s surface that is covered with water. One Proportion z – Test H0: p = 0.73 Ha: p ≠ 0.73 SRS – The data was collected randomly Appr. Normal: n (0.73) ≥ 10 AND n (1 – (0.73)) ≥ 10 Population of potential volleys is ≥ 10 · n Since the P-Value is NOT less than α = 0.05 we can NOT REJECT H0. There is NO evidence to support that proportion of Earth covered by water is NOT equal to 73% Since the P-Value is less than α = 0.05 we REJECT H0 . There is sufficient evidence that the proportion of Earth covered by water is not 71%.

HW: Page 471: 13, 14 H0: p = 0.34 Ha: p ≠ 0.34 p = True prop. of student with perfect attendance during one month SRS – The data was collected randomly Appr. Normal: 8302 (0.34) ≥ 10 AND 8302 (1 – (0.34)) ≥ 10 Population ≥ 10 · n

p = The true proportion of auto accidents involving teenagers. b.) An insurance company checks police records on 582 accidents selected at random. Teenagers were involved in 91 of them. A Politician interested in raising the drivers licenses age claims that one out of every five (p = .20) auto accidents is due to teenagers. Is there supporting evidence that the proportion is different than this? p = The true proportion of auto accidents involving teenagers. One Proportion z – Test H0: p = 0.2 Ha: p ≠ 0.2 1. SRS – Stated 2. Population of National Auto Accidents ≥ 10(582) 3. 582(.2) ≥ 10 582(1 - .2) ≥ 10 Since the P-Value is less than α = 0.05 the data IS significant . There is strong evidence to REJECT H0 . The Politician is wrong.

2. Population of AP Stat students ≥ 10(16) NO!!!! WARM –UP EXAMPLE: You have learned that the 90% of the students in AP Statistics are earning an “A”. You don’t believe it! You gather a sample of 16 of your friends and find that only 10 have “A” averages. What is the probablilty that 10/16 or less have “A” averages if 90% really do? Write the Hypothesis first! H0: p = 0.90 Ha: p < 0.90 P(p < 10/16) 1. SRS – NO!!!!!!!!!!!!!!! 2. Population of AP Stat students ≥ 10(16) NO!!!! 3. 16(.9) ≥ 10 16(1 - .9) ≥ 10 NO!!!!!!!!!!!!!1

WARM - UP A company wants to estimate the true proportion of consumers that may buy the company’s newest product. What sample size must be collected in order to estimate this proportion within 3% points with 90% Confidence? Since no p was given, use the conservative p* = .5

One Proportion Significance Test Is the die a Fair Die? p = The true proportion of times the die lands on “1”. One Proportion z – Test H0: p = .1667 Ha: p ≠ .1667 SRS – The data was collected randomly Population of potential rolls ≥ 10 · n n · (.1667) ≥ 10 AND n · (1 – (.1667)) ≥ 10 Since the P-Value is NOT less than α = 0.05 the data IS NOT significant . There is NO evidence to REJECT H0 . The die Not weighted and is Fair. Since the P-Value is less than α = 0.05 the data IS significant . There is strong evidence to REJECT H0 . The die is weighted and thus unfair.