1 / 22

6.5 One and Two sample Inference for Proportions

6.5 One and Two sample Inference for Proportions. np >5; n(1-p)>5 n independent trials; X=# of successes p=probability of a success Estimate:. Mean and variance of.

satya
Download Presentation

6.5 One and Two sample Inference for Proportions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 6.5 One and Two sample Inference for Proportions • np>5; n(1-p)>5 • n independent trials; • X=# of successes • p=probability of a success • Estimate:

  2. Mean and variance of When n is large, approximate probabilities for can be found using the normal distribution with the same mean and standard deviation.

  3. An approximate confidence interval for p is

  4. Sample Size • The sample size required to have a certain probability that our error (plus or minus part of the CI) is no more than size ∆ is

  5. If you know p is somewhere … • If then maximum p(1-p)=0.3(1-0.3)=0.21 • If then maximum p(1-p)=0.4(1-0.4)=0.24

  6. Estimate p(1-p) by substitute p with the value closest to 0.5 (0, 0.1), p=0.1 (0.3, 0.4), p=0.4 (0.6, 1.0), p=0.6

  7. Example • A state highway dept wants to estimate what proportion of all trucks operating between two cities carry too heavy a load • 95% probability to assert that the error is no more than 0.04 • Sample size needed if • p between 0.10 to 0.25 • no idea what p is

  8. Solution • ∆=0.04, p=0.25 Round up to get n=451 • ∆=0.04, p(1-p)=1/4 n=601

  9. Tests of Hypotheses • Null H0: p=p0 • Possible Alternatives: HA: p<p0 HA: p>p0 HA: pp0

  10. Test Statistics • Under H0, p=p0, and • Statistic: is approximately standard normal under H0 . Reject H0 if z is too far from 0 in either direction.

  11. Rejection Regions

  12. Equivalent Form:

  13. Example • H0: p=0.75 vs HA: p0.75 • =0.05 • n=300 • x=206 • Reject H0 if z<-1.96 or z>1.96

  14. Observed z value • Conclusion: reject H0 since z<-1.96 • P(z<-2.5 or z>2.5)=0.0124<a reject H0.

  15. Example • Toss a coin 100 times and you get 45 heads • Estimate p=probability of getting a head Is the coin balanced one? a=0.05 Solution: H0: p=0.50 vs HA: p0.50

  16. Enough Evidence to Reject H0? • Critical value z0.025=1.96 • Reject H0 if z>1.96 or z<-1.96 • Conclusion: accept H0

  17. Truth = surgical biopsy Cancer Present Cancer Absent Total Result Positive 140 80 220 FNA status Results Negative 10 910 920 150 990 1140 Total Another example • The following table is for a certain screening test

  18. Test to see if the sensitivity of the screening test is less than 97%. • Hypothesis • Test statistic

  19. What is the conclusion? • Check p-value when z=-2.6325, p-value = 0.004 • Conclusion: we can reject the null hypothesis at level 0.05.

  20. Truth = surgical biopsy Cancer Present Cancer Absent Total Result Positive 14 8 22 FNA status Results Negative 1 91 92 114 15 99 Total One word of caution about sample size: • If we decrease the sample size by a factor of 10,

  21. And if we try to use the z-test, P-value is greater than 0.05 for sure (p=0.2026). So we cannot reach the same conclusion. And this is wrong!

  22. So for test concerning proportions We want np>5; n(1-p)>5

More Related