Section 9.2 – Sample Proportions Chapter 9 Section 9.2 – Sample Proportions

The Sampling Distribution of 𝑝 As you learned in section 9.1, in order to estimate a population parameter, p, you use a sample statistic, 𝑝 . Recall that 𝑝 is the sample proportion of successes: 𝑝 = count of "successes" in sample size of sample = 𝑋 𝑛 Since the values of X and 𝑝 will vary with repeated samples, both X and 𝑝 are random variables. Provided that the population is much larger than the sample (by at least 10 times), the count X will follow a binomial distribution.

Sampling Distribution of a Sample Proportion Choose an SRS of size n from a large population with population proportion p having some characteristic of interest. Let 𝑝 be the proportion of the sample having that characteristic. Then: The mean of the sampling distribution is exactly p. 𝜇 𝑝 =𝑝 The standard deviation of the sampling distribution is: 𝜎 𝑝 = 𝑝(1−𝑝) 𝑛 Rule of thumb #1: In order to use the standard deviation of 𝑝 , the population must be at least 10 times as large as the sample.

Using the Normal Approximation for 𝑝 Recall from section 9.1 that the shape of the sampling distribution of 𝑝 is approximately normal and is closer to normal when the sample size n is larger. Rule of thumb #2: We will use the normal approximation to the sampling distribution of 𝑝 for values of n and p that satisfy 𝑛𝑝≥10 and 𝑛(1−𝑝)≥10

Example 9.7 – Applying to College A polling organization asks an SRS of 1,500 first-year college students whether they applied for admission to any other college. In fact, 35% of all first-year students applied to colleges besides the one they are attending. What is the probability that the random sample of 1,500 students will give a result within 2 percentage points of this true value? First check rule of thumb #1 & #2: The population is all first year college students in the U.S. and since the sample size is 1500, the population would have to contain at least 10(1500) =15,000 first year college students in order to use the standard deviation formula. Since there are over 1.7 million first year college students we can proceed. We can use the normal approximation since np = 1500(.35) 525 and n(1-p) = 1500(.65) = 975 are both greater than 10. The normal approximation to the sampling distribution of p.

Example 9.7 – Applying to College We want to find the probability that 𝑝 falls between .33 and .37. 𝑧= 𝑝 −𝑝 𝜎 𝑝 = .33−.35 (.35)(1−.35) 1500 =−1.63 𝑧= 𝑝 −𝑝 𝜎 𝑝 = .37−.35 (.35)(1−.35) 1500 =1.63 𝑃 0.33≤ 𝑝 ≤0.37 =𝑃 −1.63≤𝑧≤1.63 =.9484−.0516 = .8968 This means that almost 90% of all samples will give a result within 2 percentage points of the truth about the population.

Example 9.8 – Survey Undercoverage One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample is to compare the sample with known facts about the population. About 11% of American adults are black. The proportion p of blacks in a SRS of 1,500 adults should therefore be close to .11. It is unlikely to be exactly .11 because of sampling variability. If a national sample contains only 9.2 % blacks, should we suspect that the sampling procedure is somehow underrepresenting blacks? Test both rule of thumbs: The population of all black American adults is larger than 10(1500) = 15,000 so the standard deviation formula can be used A normal approximation can be used since np = 1500(.11) = 165 and n(1-p) = 1500(.89) = 1335.

Example 9.8 – Survey Undercoverage 𝑧= 𝑝 −𝑝 𝜎 𝑝 = .092−.11 (.11)(1−.11) 1500 =−2.23 𝑃 𝑝 ≤0.092 =𝑃 𝑧≤−2.23 =.0129 Only 1.29% of all samples would have so few blacks which would give good reason to believe that the sampling procedure underrepresented blacks.

Here is a video lesson if you would like to see more examples: Homework: p.511 #’s 19-21, 23, 25, 26, & 30 Here is a video lesson if you would like to see more examples: http://www.youtube.com/watch?v=xZx6wKUEShs