Tests for Population Proportion(s)

Transcription

1 Tests for Population Proportion(s) Esra Akdeniz April 6th, 2016

2 Motivation We are interested in estimating the prevalence rate of breast cancer among 50- to 54-year-old women whose mothers have had breast cancer. Suppose that in a random sample of 10,000 such women, 400 are found to have had breast cancer at some point in their lives. The best point estimate of the prevalence rate p is the sample proportion p = = 0.04 Given large studies, assume the prevalence rate of breast cancer for U.S. women in this age group is about 2%. The question is: How compatible is the sample rate of 4% with a population rate of 2%?

3 Motivation Another way of asking this question is to restate it in terms of hypothesis testing: p = prevalence rate of breast cancer in 50- to 54-year-old women whose mothers have had breast cancer H 0 : p = 0.02 = p 0 vs. H 1 : p 0.02 How do we test this hypothesis?

4 Introduction So far, we dealt with hypothesis concerning population mean. In this lecture, we will learn how to make inference on sample proportions: proportion of times an event occurs rather than the number of times. Example: Every year, approximately 3500 babies were delivered in the Vienna General Hospital in the mid-nineteenth century. Approximately 500 women develop puerperal fever an infection developing during childbirth. Assume X Bin(n, p). Normal approximation to binomial distribution: IF np 5 AND n(1 p) 5, then under H 0, p is approximately normally distributed.

5 Test about a population proportion Let p denote the proportion of individuals or objects in a population who possess a specified property (labeled as S ). Let X be the number of Ss in the sample. Then p = X n is the sample proportion. X is a binomial random variable with parameters p and n, i.e. X Bin(n, p). Furthermore, when the sample size n itself is large, both X and p are approximately normally distributed, i.e. X N(np, np(1 p)) and p N(p, p(1 p) n ). Test about population proportion p will depend on the sample size.

6 Test about a population proportion Large-Sample Tests When the sample size is large (n 30) (IF np 5 AND n(1 p) 5), p is approximately normally distributed with mean p and variance p(1-p)/n. In particular, under the null hypothesis H 0 : p = p 0, p is approximately normally distributed with mean p 0 and variance p 0(1 p 0)/n, i.e. p N(p 0, p 0(1 p 0)/n). Therefore the test statistic Z = p p 0 p0(1 p 0)/n has approximately a standard normal distribution.

7 Example For breast cancer example we compute the test statistic z = = p p 0 p0(1 p 0)/n = (0.98)/ = 14.3 z 1 α/2 = z = 1.96 Since 14.3 > 1.96, H 0 is rejected using two sided test with α = 0.05.

8 Figure: Acceptance and rejection regions for the one-sample binomial test, normal-theory method (two-sided alternative)

9 Example Suppose that we select a random sample of 30 individuals from the population of adults in Turkey. Assume that the probability that a member of this population currently smokes cigarettes, cigars or pipes is equal to Therefore, the total number of smokers in the sample is binomial with n = and p =. For this sample, what is the probability that six or fewer of its members smoke? Continuity correction.

10 Sample proportion and its distribution Sample proportion is denoted as ˆp and ˆq = 1 ˆp. Its distribution is normal with mean= and variance =, using CLT. Confidence interval for p: ( p z 1 α/2 p q/n, p + z1 α/2 p q/n )

11 Hypothesis Testing Null hypothesis: H 0 : p = p 0 OR H 0 : p p 0 OR H 0 : p p 0. Test statistic: z = ˆp p 0 p0 (1 p 0 ) n

12 Example Consider the distribution of five-year survival for individuals under 40 who have been diagnosed with lung cancer. This distribution has an unknown population mean p. In a randomly selected sample of 52 patients, only six survive five-years. Compute the sample proportion. Find the 95% confidence interval for the population proportion. Test the hypothesis that the population proportion is equal to at significance level 0.05.

13 Comparison of two Population Proportions Assume X Bin(m, p 1), Y Bin(n, p 2) and they are independent. Normal approximation under conditions: m ˆp 1 5, m(1 ˆp 1) 5, n ˆp 2 5, n(1 ˆp 2) 5. Confidence interval: ( p 1 p 2 ± z 1 α/2 p1(1 p 1) m + ) p2(1 p2) n

14 Hypothesis Testing Null hypothesis: H 0 : p 1 p 2 = 0 Test statistic: z = ˆp 1 ˆp 2 0 ˆp(1 ˆp) m + ˆp(1 ˆp) n N(0, 1) Under H 0 the proportions are assumed equal therefore a common p value is estimated by ˆp = x1 + x2 m + n

15 Example In a study investigating morbidity and mortality among pediatric victims of motor vehicle accidents, information regarding effectiveness of seat belts was collected over an 18-month period. Two random samples were selected, one from the population of children who were wearing a seat belt at the time of the accident, and the other from the population the population who were not. In the sample of 123 children who were wearing a seat belt at the time of the accident, 3 died. In the sample of 290 children who were not wearing a seat belt, 13 died. We want to test whether the population proportions of these two populations are equal at significance level 0.05.