News reports tell us that the average American is overweight. Many of us have tried to trim down to our weight when we finished high school or college. And, in fact, only 20% of adults say they do not suffer from weight-loss woes. Suppose that the 20% figure is correct, and that a random sample of n = 120 adults is selected.
a. Does the distribution of p, the sample proportion of adults who do not suffer from excess weight, have an approximate normal distribution? If so, what is its mean and standard deviation?
b. What is the probability that the sample proportion p exceeds .25?
c. What is the probability that p lies within the interval .25 to 30?
d. What might you conclude about p if the sample proportion exceeded .30?
To answer these questions, we can use the properties of the sample proportion and the Central Limit Theorem. The distribution of the sample proportion is approximately normal when the sample size is large enough (n > 30), and the population from which the sample is drawn is not extremely skewed.
a. Yes, the distribution of p has an approximate normal distribution. The mean of the sample proportion, denoted as μp, is equal to the population proportion, which is given as 0.20 in this case. The standard deviation of the sample proportion, denoted as σp, is calculated as the square root of (p(1-p)/n), where p is the population proportion and n is the sample size. Substituting the given values, we have σp = sqrt(0.20(1-0.20)/120) ≈ 0.0314.
b. To find the probability that the sample proportion p exceeds 0.25, we can standardize the distribution using the Z-score formula: Z = (p - μp) / σp. The Z-score can be calculated as (0.25 - 0.20) / 0.0314 ≈ 1.59. We can then look up the corresponding probability from the standard normal distribution table (or use statistical software) and find the area under the curve to the right of Z = 1.59. This will give us the probability that p exceeds 0.25.
c. To find the probability that p lies within the interval 0.25 to 0.30, we need to find the areas under the curve to the left and right of these values. We can calculate the corresponding Z-scores for each end of the interval: Z1 = (0.25 - 0.20) / 0.0314 ≈ 1.59 and Z2 = (0.30 - 0.20) / 0.0314 ≈ 3.18. Then, using the standard normal distribution table, we can find the probability associated with Z1 and subtract it from the probability associated with Z2 to get the desired probability.
d. If the sample proportion p exceeds 0.30, we can conclude that a larger percentage of the population does not suffer from excess weight than initially estimated (20%). However, it is important to note that this conclusion is based on a sample, not the entire population. In order to make more generalizable conclusions, it would be necessary to conduct a larger study or evaluate additional information.