Chapter 20 and 21 combined testing hypotheses about proportions 2013

20. Nov 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
Chapter 20 and 21 combined testing hypotheses about proportions 2013
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Chapter 20 and 21 combined testing hypotheses about proportions 2013

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  1. The null hypothesis is H0: p=0.75., HA: p>0.75With a P-value of 0.0001, this is very strong evidence against the null hypothesis. We can reject H0 and conclude that the improved version of the drug gives relief to a higher proportion of patients.
  2. 4. The parameter of interest is the proportion, p, of all delinquent customers who will pay their bills. H0: p = 0.30 and HA: p> 0.30.
  3. 5. The very low P-value leads us to reject the null hypothesis. There is strong evidence that the video tape is more effective in getting people to start paying their debts than just sending a letter had been.
  4. P = proportion of students in districts like this one who drop out.H0: p = 10.3 (or p<=.103)HA: p >.103Phat= .118P0= .103Q0= .897Sd= sqrt((.103)(.897)/1782)= .007Z= (.118-.103)/.007=2.14So Normalcdf(2.14, 99)= .016=1.6%= p-valueThis p value is really low so we are going to reject the null hypothesis and conclude that ______________.
  5. P = proportion of car crashes occurring within 5 miles of home.H0: p = 50 (or p<=.50)HA: p >.50STAT, TESTS #5, enter in infoZ= 4.195P = .00000136This p value is extremely small so we reject the null hypothesis and conclude …
  6. a) Independence assumption: The Euro spins are independent. One spin is not going to effect the others. (With true independence, it doesn’t make sense to try to check the randomization condition or the 10% condition. These verify our assumption of independence, and we don’t need to do that!)Success/Failure condition: npˆ = 140 and nqˆ = 110 are both greater than 10, so the sample is large enough.Since the conditions are met, we can use a one-proportion z-interval to estimate the proportion of heads in Euro spins.We are 95% confident that the true proportion of heads when a Euro is spun is between 0.498 and 0.622.b) Since 0.50 is within the interval, there is no evidence that the coin in unfair. 50% is a plausible value for the true proportion of heads. (That having been said, I’d want to spin this coin a few hundred more times. It’s close!)c) The significance level is α = 0.05. It’s a two-tail test based on a 95% confidence interval.
  7. 7. With a z-score of 0.62, you can’t reject the null hypothesis. The experiment shows no evidence that the wheel is not fair.8. At alpha=0.05, you can’t reject the null hypothesis because 0.30 is contained in the 90% confidence interval- it’s plausible that sending the DVDs is no more effective than just sending letters.9. The confidence interval is from 29% to 45%. The DVD strategy is more expensive and may not be worth it. We can’t distinguish the success rate from 30% given the results of this experiment, but 45% would represent a large improvement. The bank should consider another trial, increasing their sample size to get a narrower confidence interval.
  8. 10. A Type I error would mean deciding that the DVD success rate is higher than 30% when it really isn’t. They would adopt a more expensive method for collecting payments that’s no better than the less expensive strategy.11. A Type II error would mean deciding that there’s not enough evidence to say that the DVD strategy works when in fact it does. The bank would fail to discover an effective method for increasing their revenue from delinquent accounts.
  9. a) Type II. The filter decided that the message was safe, when in fact it was spam.b) Type I. The filter decided that the message was spam, when in fact it was not.c) This is analogous to lowering alpha. It takes more evidence to classify a message as spam.