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Section 7.2 Sample Proportions ˆ The Sampling Distribution of p How good is the statistic pˆ as an estimate of the parameter p? The sampling distribution of pˆ answers this question. Consider the approximate sampling distributions generated by a simulation in which SRSs of Reese’s Pieces are drawn from a population whose proportion of orange candies is 0.45. What do you notice about the shape, center, and spread of each? ˆ The Sampling Distribution of p What did you notice about the shape, center, and spread of each sampling distribution? Shape : In some cases, the sampling distribution of pˆ can be approximated by a Normal curve. This seems to depend on both the sample size n and the population proportion p. Center : The mean of the distribution is m pˆ = p. This makes sense because the sample proportion pˆ is an unbiased estimator of p. Spread : For a specific value of p , the standard deviation s pˆ gets smaller as n gets larger. The value of s pˆ depends on both n and p. There is an important connection between the sample proportion p̂ and the number of "successes" X in the sample. pˆ = count of successes in sample X = size of sample n ˆ The Sampling Distribution of p In Chapter 6, we learned that the mean and standard deviation of a binomial random variable X are mX = np sX = np(1 - p) Since pˆ = X /n = (1/n) × X, we are just multiplying the random variable X by a constant (1/n) to get the random variable pˆ . Therefore, 1 m pˆ = (np) = p n pˆ is an unbiased estimator or p The Sampling Distribution of p̂ The standard deviation of the sampling distribution is: 1 np(1 - p) s pˆ = np(1 - p) = = 2 n n p(1 - p) n As sample size increases, the spread decreases. As long as the 10% condition is satisfied: n < (1/10)N or 10n < N As n increases, the sampling distribution becomes approximately Normal. Before you perform Normal calculations, check that the Normal condition is satisfied: np ≥ 10 and n(1 – p) ≥ 10. ˆ The Sampling Distribution of p An SRS of 1500 first-year college students were asked whether they applied for admission to any other college. In fact, 35% of all first-year students applied to colleges beside the one they are attending. What is the probability that the poll will be within 2 percentage points of the true p? Step 1: State the distribution and the values of interest. 𝜇𝑝 = 𝑝 = 0.35 𝑃 0.33 < 𝑝 < 0.37 = ? Check conditions!!!!! 10n < N np ≥ 10 and n(1 – p) ≥ 10. 10(1500) = 15,000 < population of all first year college students 1500(.35) = 525 ≥ 10 and 1500(.65) = 975 ≥ 10 Since we meet our conditions we can consider the distribution of 𝑝 to be approximately Normal. Step 2: Perform calculationsshow your work!!!!! 𝜇𝑝 = 𝑝 = 0.35 (0.35)(0.65) = .0123 pˆ 1500 Normalcdf (lwr: 0.33, upr: 0.37, μ: 0.35, : 0.0123) = 0.896 𝑃 0.33 < 𝑝 < 0.37 = 0.896 Step 3: About 90% of all college student SRSs of size 1500 fall within 2% of the real p of students who applied to other schools . Planning for college The superintendent of a large school district wants to know what proportion of middle school students in her district are planning to attend a four-year college or university. Suppose that 80% of all middle school students in her district are planning to attend a four-year college or university. What is the probability that an SRS of size 125 will give a result within 7 percentage points of the true value? Solution: Step 1: State the distribution and the values of interest. We want to find the probability that the percentage of middle school students who plan to attend a fouryear college or university falls between 73% and 87%. Converting to proportions, we want to find, P(0.73 ≤ 𝑝 ≤ 0.87). The mean of the sampling distribution of 𝑝 is 𝜇𝑝 = 𝑝= 0.80. Step 2: Check Conditions Because the school district is large, we can assume that there are more than 10(125) = 1250 middle school students. We can consider the distribution of 𝑝 to be approximately Normal because np = 125(0.80) = 100 > 10 and n(1 – p) = 125(0.20) = 25 > 10 0.80(0.20) 0.036 Step 3: Answer the question. p̂ 125 The standardized scores for the boundary values are z 0.73 0.80 1.94 0.036 z 0.87 0.80 1.94 0.036 P(−1.94 ≤ 𝑧≤ 1.94)= 0.9476 Using technology: The command normalcdf(lower: 0.73, upper: 0.87, μ: 0.80, : 0.036) gives and area of 0.9482. Answer in Context!!! About 95% of all SRSs of size 125 will give a sample proportion within 7 percentage points of the true proportion of middle school students who are planning to attend a four-year college or university. The Gallup Poll asked a random sample of 1785 adults whether they attended church during the past week. Let 𝑝 be the proportion of people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. a. Describe the sampling distribution of 𝑝. b. Of the poll respondents, 44% said they did attend church last week. Find the probability of obtaining a sample of 1785 adults in which 44% or more say they attended church last week if the newspaper report’s claim is true. Does this poll give convincing evidence against the claim? Explain. c. What sample size would be required to reduce the standard deviation of the sampling distribution to one-third the value you found? Justify your answer. Have you ever noticed that some goldfish crackers have smiles and some don’t? We are going to estimate the proportion of all goldfish crackers that have faces. Select a Dixie cup of cheddar Goldfish crackers. Don’t eat any yet!