People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior.

a. Show the sampling distribution of ( ), the proportion of groceries thrown out by your sample respondents (to 4 decimals)?

b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion (to 4 decimals)?

c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion (to 4 decimals)?

To answer these questions, we need to use the concept of sampling distribution. The sampling distribution refers to the distribution of sample statistics (in this case, the sample proportion) based on repeated sampling from the same population.

a. Show the sampling distribution of ( ), the proportion of groceries thrown out by your sample respondents (to 4 decimals)?

The sampling distribution of the proportion of groceries thrown out by the sample respondents can be approximated by a normal distribution if the sample size is sufficiently large. Given that the population proportion is 12%, denoted as p.

Therefore, the mean of the sampling distribution is the same as the population proportion: μ = p = 0.12.

The standard error (SE) of the sampling distribution is calculated using the formula: SE = sqrt(p * (1 - p) / n), where n is the sample size.

In this case, n = 540, so SE = sqrt(0.12 * (1 - 0.12) / 540) = 0.0146 (rounded to 4 decimals).

Assuming the sampling distribution is approximately normal, the sampling distribution of the proportion of groceries thrown out by the sample respondents can be represented as N(0.12, 0.0146) (mean, standard error).

b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion (to 4 decimals)?

To find the probability that the sample proportion is within ±.03 of the population proportion, we need to find the area under the normal curve within that range.

Using z-scores, we can standardize the range (.03) using the formula: z = (x - μ) / SE, where x is the value we are interested in, μ is the population proportion, and SE is the standard error.

For the lower limit, z = (.12 - .03) / 0.0146 = 6.16 (rounded to 2 decimals).
For the upper limit, z = (.12 + .03) / 0.0146 = 11.64 (rounded to 2 decimals).

Using a normal distribution table or statistical software, we can find the probability associated with these z-scores. The probability of the sample proportion being within ±.03 of the population proportion is the cumulative probability between these two z-scores.

c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion (to 4 decimals)?

Similarly, we can use the same steps as in part b to find the probability of the sample proportion being within ±.015 of the population proportion. We need to calculate the z-scores for the lower and upper limits, and then use a normal distribution table or statistical software to find the cumulative probability between those z-scores.