Introductory Statistics, Lecture 18

Has the percentage of overweight adults changed over recent years? Math 140 Introductory Statistics „ Health experts agree that one of the best ways to improve health and extend life among U.S. residents is to reduce the percentage of those overweight. What is not so clear is whether the percentage of overweight adults has changed over recent years. It may seem that more are overweight simply because the issue has garnered much media attention. On the other hand, with the current emphasis on exercise and healthy eating, perhaps the percentage of overweight adults has actually decreased. So the question for investigation is whether the proportion of overweight adults has changed over the past 5 years. Lecture 18 1 2 Has the percentage of overweight adults changed over recent years? Inference for Experiments „ One source of data on this issue is the Harris „ The significance test follows along the same lines as for surveys. The main difference is that instead of requiring the samples to be random and independent, we require the subjects of the experiment to be randomly assigned to their treatments. „ If comparisons are to be made on subjects or units without random assignment, such as in comparing cancer rates in a group of smokers with rates in a group of nonsmokers, the investigation is called an observational study. poll on lifestyles conducted every year with a random sample of about 1000 adults. In 2004, 30% of adults in the sample were 20% or more overweight, while in 2009 the percentage was 32%. Does this represent a real change in population proportions at the 10% level of significance? [Source: http://www.harrisinteractive.com/.] 3 4 Protection from the Flu Protection from the Flu „ For use during influenza epidemics, two types „ During influenza season, 446 nurses who were caring for patients with respiratory illness including fever were randomly divided into two groups. One group of 225 nurses received surgical masks and 221 received fitted N95 respirators. The display below gives the number of nurses in each group who got influenza. Is this statistically significant evidence that one type of respiratory protection is better than another? of respiratory protection are available: surgical masks and N95 respirators. Surgical masks are less expensive than N95 respirators, and in greater supply, but they do not fit tightly and do not protect wearers from smaller particles. An experiment was performed to see which type of mask was more effective. 5 6 Notation:Proportions, Sample Proportions, and Sample Sizes. Two AIDS treatments „ Consider a clinical trial experiment comparing two treatments for AIDS-related complex (ARC). The investigators want to find out if there is a significant difference on the survival rates of patients who had already developed AIDS. They have two treatments, patients who were given AZT and patients who were given AZT + ACV. Here’s the data n1 = 69 sample size of patients treated with AZT n2 = 62 sample size of patients treated with AZT+ACV pˆ 1 = Treated with Survived? AZT AZT + ACV Total No 28 13 41 Yes 41 49 90 Total 69 62 131 pˆ 2 = 7 41 69 49 62 proportion of patients in the sample treated with AZT that survived proportion of patients in the sample treated with AZT+ACV that survived p1 True proportion of survival if all patients were treated with AZT (unknown) p2 True proportion of survival if all patients were treated with AZT+ACV (unknown) 8 Assumptions about the difference of two proportions. Checking the conditions „ If we have obtain two independent sample proportions „ Check the conditions. „ We assume that subjects were randomly assigned to treatments. „ All of the following are greater than 5 p p̂1 and p̂p2 , then the distribution of the difference of the pˆ 1 − pˆ 2 is approximately normal as two proportions p1-p2 long as each proportion satisfies the following three conditions: p̂1 and p̂p2 is a simple random sample from a „ Each sample p „ „ n1pˆ 1 = 69(41/ 69) = 41, n2pˆ 2 = 62(49/ 62) = 49 n1(1− pˆ 1) = 69(1− 41/ 69) = 28 binomial population, and they are independent from each other. or, in case of experiments, subjects were randomly assigned to their treatments. ˆ11,,nn12ppˆ2,2,nn11(1(1–− ppˆ11),),n − pˆ 2 ) All the numbers nn11pp and 2 (1n are at least 5. Each of the two population sizes is at least 10 times the sample size. n2 (1− pˆ 2 ) = 62(1− 49/ 62) =13 „ The population size of AIDS patients that could potentially be treated with AZT is clearly greater than 10 times 69 = 690 The population size of AIDS patients that could potentially be treated with AZT+ACV is clearly greater than 10 times 62 = 620 9 10 Writing the hypothesis The Test Statistic „ The null hypothesis „ H0 : The new therapy (AZ+ACV) is as good as the old therapy (AZT). That is the survival rate if all patients were treated with AZT+ACV would be equal to the survival rate if all patients were treated with AZT. In symbols p 1 = p 2 or p 1 − p 2 = 0 „ The test statistic in general follows the form estimate − parameter Test Statistic = standard deviation of estimate „ Now, the standard deviation of the estimate under the assumption that p1 = p2 can be approximated by „ The parameter is p1 – p2, and if the null hypothesis is true then p1 – p2 = 0. „ The estimate is what we obtain for the difference of the two proportions according to our samples. „ The alternate hypothesis „ Ha : The new therapy (AZ+ACV) is better than the old therapy (AZT). That is the survival rate if all patients were treated with AZT+ACV would be greater than the survival rate if all patients were treated with AZT. In symbols p 1 < p 2 or p 1 − p 2 < 0  1 1   + pˆ (1 − pˆ )  n n 2   1 „ Where p p̂ is the pooled estimate pˆ = 90 total survived = ≈ 0 .687 131 grand total estimate = pˆ 1 − pˆ 2 = 0 .594 − 0 .790 = − 0 .196 11 12 The Test Statistic Test Statistic = P-Value estimate − parameter „ standard deviation of estimate „ The parameter is p1 – p2 = 0, estimate = pˆ 1 − pˆ 2 = 0 .594 − 0 .790 = − 0 .196 „ The standard deviation of the estimate  1 1 pˆ (1 − pˆ ) +  n1 n 2     pp̂ is the pooled estimate pˆ = „ To get the P-Value we use the fact that the distribution of the difference is approximately normal. total survived 90 = ≈ 0 .687 grand total 131 „ so the standard deviation of the estimate is: 1   1 ( 0 .687 )(1 − 0 .687 ) +   69 62  = 0 .081145 „ and the test statistic: „ Since our test is one-sided we need to calculate the probability that the difference is less than our teststatistic. „ We can find P by doing normalcdf(– 999999,– 2.4154)=.007858 − 0 .196 − 0 = − 2 .415429 0 .081145 13 Conclusion „ Since the P-Value equals 0.78% and this is definitely less than 5% we reject the null hypothesis. „ If both treatments were equally effective (null hypothesis is true) then there is only a 0.78% pˆ 1 − pˆ 2 as small or chance of getting a difference p1-p2 smaller than – 0.196. This probability is so small that we are confident that if all subjects in the experiment had been given AZT+ACV there would be a larger survival rate than if they had received only AZT. 15 14