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)?
any luck
To answer these questions, we need to use the concept of sampling distribution. A sampling distribution is the distribution of a statistic (in this case, the proportion) calculated from different random samples taken from the same population.
a. The sampling distribution of the proportion (p-hat) can be approximated by a normal distribution when the sample size is large enough. In this case, we are taking a sample of 540 grocery shoppers. The formula for the standard deviation of the sampling distribution of the proportion is:
Standard Deviation = sqrt((p * (1 - p)) / n)
where p is the population proportion and n is the sample size.
Given that the population proportion is 12% (or 0.12) and the sample size is 540, we can calculate the standard deviation as follows:
Standard Deviation = sqrt((0.12 * (1 - 0.12)) / 540) = sqrt(0.1056 / 540) ≈ 0.0170
Therefore, the sampling distribution of the proportion has a standard deviation of approximately 0.0170.
b. To find the probability that the survey will provide a sample proportion within ±0.03 of the population proportion, we need to calculate the z-scores corresponding to the lower and upper bounds.
Lower Bound: p - 0.03 = 0.12 - 0.03 = 0.09
Upper Bound: p + 0.03 = 0.12 + 0.03 = 0.15
Next, we convert these values into z-scores using the formula:
z = (x - p) / Standard Deviation
Using the lower bound:
z_lower = (0.09 - 0.12) / 0.0170 ≈ -1.765
Using the upper bound:
z_upper = (0.15 - 0.12) / 0.0170 ≈ 1.765
To find the probability corresponding to these z-scores in a standard normal distribution, we can use a z-table or a statistical calculator. Looking up the z-scores in the table or using a calculator, we find that the probability of obtaining a sample proportion within ±0.03 of the population proportion is approximately 0.9241.
c. Using the same approach, to find the probability that the survey will provide a sample proportion within ±0.015 of the population proportion:
Lower Bound: p - 0.015 = 0.12 - 0.015 = 0.105
Upper Bound: p + 0.015 = 0.12 + 0.015 = 0.135
z_lower = (0.105 - 0.12) / 0.0170 ≈ -0.882
z_upper = (0.135 - 0.12) / 0.0170 ≈ 0.882
The probability of obtaining a sample proportion within ±0.015 of the population proportion is approximately 0.7704.