Question

Mrs. Bracken asks a random sample of 700 adults whether they favor giving vouchers to parents of school-age children that can be exchanged for education at any public or private school of their choice. Each school would be paid by the government on the basis of how many vouchers it collected. Suppose that 45% of the population favor this idea. (a) What is the mean of the sampling distribution of p∗, the proportion of adults in samples of size n=700 who favor giving parents of school-age children these vouchers? (b) What is the standard deviation of p? Interpret this value. (c) Check that you can use the Normal approximation for the distribution of p. (d) What is the probability that more than half of the sample are in favor? Show your work.

Answer 1

(a) The mean, often denoted by µ, of the sampling distribution of any statistic is simply the statistic calculated from the population. Here the population statistic is the proportion of adults favoring the vouchers, which is 45%. So, the mean of the sampling distribution is 0.45, or 45%.

(b) Next, we calculate the standard deviation, σ, of the sample proportion. From the central limit theorem, we know that the standard deviation for the sample proportion is given by the formula sqrt[p(1-p)/n], where p is the population proportion and n is the sample size. Here, p = 0.45 and n = 700. Plugging these values into the formula, we get the standard deviation as 0.0188. This value demonstrates the amount of variation or dispersion of the set of sample proportions from the mean.

(c) The normal approximation can be used for the distribution of the sample proportion when the conditions np > 5 and n(1-p) > 5 are met, where n is the sample size and p is the population proportion. In this case, n = 700 and p = 0.45. Calculating np gives us 315 and n(1-p) gives us 385. Since both of these values are greater than 5, we can use the Normal approximation for the distribution of the sample proportion.

(d) To find the probability that more than half of the sample favors the vouchers, we are interested in finding P(p > 0.5). To do this, we first need to compute the z-score for 0.5. The z-score is given by the formula z = (x - µ) / σ. Here, x = 0.5, µ is the mean of the sampling distribution (which is 0.45), and σ is the standard deviation (which is 0.0188). Plugging these into the formula, we get a z-score.

Then, to find the actual probability, we use the standard normal table (also known as the Z-table), which tells us the area to the left of any given z-score. But since we want the area to the right of the z-score (i.e. the probability that more than half the sample favors the vouchers), we need to subtract the table value from 1. Doing so gives us the probability as approximately 0.0039, or 0.39%.

This indicates a very slim likelihood that more than 50% of a sample of 700 adults pulled randomly from our defined population would favor this school voucher proposition.