One sample statistical tests: Sample Proportion

1. One sample statistical tests:Sample Proportion Math 137 Fresno State Burger

2. The normal approximation to the binomial… Statistics for proportions are based on a normal distribution, because the binomial can be approximated as ‘normal’ if and

3. Differs by a factor of n. Differs by a factor of n. P-hat stands for “sample proportion.” Recall: stats for proportions For binomial: For proportion:

4. BUT… if you knew p you wouldn’t be doing the experiment! Sampling distribution of a sample proportion p=true population proportion. Always a normal distribution!

5. Practice Problem A. A fellow researcher claims that at least 15% of smokers fail to eat any fruits and vegetables at least 3 days a week. You find this hard to believe and decide to check the validity of this statistic by taking a random (representative) sample of smokers. Do you have sufficient evidence to reject your colleague’s claim if you discover that 17 of the 200 smokers in your sample eat no fruits and vegetables at least 3 days a week?

6. Answer 1. What is your null hypothesis? Null hypothesis: p=proportion of smokers who skip fruits and veggies frequently >= .15 Alternative hypothesis: p < .15 2. What is your null distribution? Var( ) = .15*.85/200 = .00064 SD( ) = .025 ~ N (.15, .025) 3. Empirical evidence: 1 random sample: = 17/200 = .085 4. Z = (.085-.15)/.025 = -2.6 p-value = P(Z<-2.6) = .0047 5. Sufficient evidence to reject the claim.

7. Corresponding confidence intervals… For a mean: For a proportion: Tn-1 approaches Z for large n. ** If np (expected value)<5, use exact binomial rather than Z approximation…

8. Symbol overload! n: Sample size Z: Z-statistic (standard normal) p: (“p-hat”): sample proportion X: (“X-bar”): sample mean s:Sample standard deviation p0: Null hypothesis proportion

9. Practice Problem B. According to a recent poll 53% of Americans would vote for the incumbent president. If a random sample of 100 people results in 45% who would vote for the incumbent, test the claim that the actual percentage is 53%. Use a 0.10 significance level.

10. Practice Problem B. solution Fail to reject . We can not say with 90 % confidence that the true proportion of voters who would vote for the incumbent is not equal to 53%.

11. Practice Problem C. In a clinical study of an allergy drug, 108 of the 203 subjects reported experiencing significant relief from their symptoms. At the 0.01 significance level, test the claim that more than half of all those using the drug experience relief.

12. Practice Problem C. solution Fail to reject . I can not say with 99 % confidence that the true proportion of subjects who reported experiencing significant relief from their symptoms is greater than .5

13. Practice Problem D. A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and production is not really out of control. At the 0.01 level of significance, test the manager’s claim.

14. Practice Problem D. solution Fail to reject . I can not say with 99 % confidence that the true defect rate is more than 3%. What can you say if you only need 90% confidence?