Warm-UP: A Centers for Disease Control report of randomly selected Americans, found that 411 of 1012 men and 535 of 1062 women suffered from some form.

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Warm-UP: A Centers for Disease Control report of randomly selected Americans, found that 411 of 1012 men and 535 of 1062 women suffered from some form of the common cold at least once a month. Is there evidence that this proportion of men differs from that of women? (α = 0.05) pi = The true proportion of citizens that suffer from the cold. pm = Men and pw = Women H0: pm = pw Ha: pm ≠ pw TWO Proportion z – Test 1. SRS – Both sample data were collected randomly 2. 1012· (0.41) ≥ 10 AND 1012· (1 – 0.41) ≥ 10 1062· (0.50) ≥ 10 AND 1062· (1 – 0.50) ≥ 10. 3. Population of men is ≥ 10 · (1012) Population of women is ≥ 10 · (1062) Since the P-Value is less than α = 0.05 REJECT H0 . The proportion of men who suffer from the cold does differs from that of women.

TEST IS NOW WEDNESDAY!!

Chapter 22 continuted If ZERO is NOT in the 2-Prop. C% Confidence Interval then a Ha: p1 ≠ p2 - Significance Test WILL Reject the null hypothesis [α = (1 – C)] , declaring that there is evidence of a significant difference between p1 and p2. (+#, +#) or (-#, -#) p1 – p2 ≠ 0

#3 C.I. - We are 95% Confident that the difference in the proportion of men and women who suffer from the arthritis is between 5.5% and 14%. ZERO is NOT in 95% Confidence Interval, so the α = 0.05 Significance Test WILL Reject the null hypothesis H0: pm = pw Ha: pm ≠ pw S.T. - Since the P-Value is less than α = 0.05 the data IS significant . There is STRONG evidence to REJECT H0 . The proportion of men who suffer from the cold does differs from that of women?

More with Tests #11. A study done in 2001 ask random non-smoking teenagers about their parents attitudes about smoking. Three years later of the 284 students who said their parents disapproved of it, 54 of them were now smokers. Among the 41 students who said their parents were lenient about it, 11 became smokers. Is there evidence that parental attitudes influence teenagers’ behaviors? pi = The true proportion of students who become smokers… p1 = Parents who disapprove and p2 = Parents who were lenient TWO Proportion z – Test H0: p1 = p2 Ha: p1 ≠ p2 CONDITIONS 1. SRS – Both sample data were collected randomly 2. 284 · (0.19) ≥ 10 AND 284 · (1 – 0.19) ≥ 10 41 · (0.27) ≥ 10 AND 41 · (1 – 0.27) ≥ 10

H0: p1 = p2 Ha: p1 ≠ p2 Conclusion: Since the P-Value is NOT less than α = 0.05 the data IS NOT significant . Fail to REJECT H0 . There is NO evidence to suggest that Parent attitudes affect teenage smoking behavior. e) P-value ?? f) Type I or II Error?

More Confidence Intervals z – Confidence Interval The painful wrist condition called Carpal Tunnel syndrome can be treated with surgery or wrist splints. In September 2002 a study of 176 random patients revealed that the half that had surgery, 80% showed improvement. Of the other half who use wrist splints, 54% showed improvement. Construct a 95% confidence interval for this difference. TWO Proportion z – Confidence Interval CONDITIONS 1. SRS – Both sample data were collected randomly 2. 88 · (0.8)=70.4 ≥ 10 AND 88 · (1 – 0.8)=17.6 ≥ 10 88 · (0.54)=47.52 ≥ 10 AND 88 · (1 – 0.54)=40.48 ≥ 10

Conclusion: We can be 95% Confident that the true difference in the proportion of patients who had surgery and showed improvement and the proportion of patients who wore wrist splints and showed improvement is between 0.116 and .384.

HW Page 509: 12, 14, 24 d,e H0: pN.D = pDepr. Ha: pN.D < pDepr. Obs. Study –Prospective. No Randomness P-val .013 Reject Ho. Evidence that depr. Patients are more likely to die. There is a .013 prob. that we will obs. evidence for Ha given that depression does not effect death rate. Type I H0: pN.D = pDepr. Ha: pN.D < pDepr.