Question

A local Vancouver health food store is gathering information about the income of their customers. They survey every customer and find that the average income of their customers is $50,000 / year with a standard deviation of $1,000. However, their general manager is unhappy with how expensive it was to survey every single customer. She hires you to answer the following questions:

a) If they surveyed only 20 customers, how likely would it be that the average income of the sample would be within $500 of the actual average income among the population (which they now know to be 50,000)? (3 points)

b) If they surveyed 50 customers, how likely would it be that the average income of the sample would be within $500 of the actual average income among the population? (3 points)

c) Explain the difference between your answers to a) and b). (3 points)

d) The general manager asks you to suggest a suitable sample size. What does your answer depend on? (3 points)

Answer 1

a) The probability that the average income of a sample of 20 customers is within $500 of the actual average income is approximately 98.57%. b) The probability that the average income of a sample of 50 customers is within $500 of the actual average income is approximately 99.98%. c) The difference between the answers to parts a) and b) is due to the larger sample size in part b), which increases the likelihood of the sample average being within a certain range of the population average. d) The suitable sample size depends on the desired level of certainty, margin of error, and variability of the population.

Step-by-step explanation:

a) To find the probability that the average income of a sample of 20 customers is within $500 of the actual average income, we can use the Central Limit Theorem. Since the population standard deviation is known to be $1,000, we can calculate the standard error as $1,000/sqrt(20) = $223.61. Then, we can calculate the z-score for a $500 margin as (500-0)/223.61 = 2.237. Using a standard normal distribution table, we can find that the probability of a z-score less than 2.237 is approximately 0.9857. Therefore, the likelihood that the average income of the sample would be within $500 of the actual average income is approximately 0.9857 or 98.57%.

b) To find the probability that the average income of a sample of 50 customers is within $500 of the actual average income, we follow the same steps as in part a). The standard error would be $1,000/sqrt(50) = $141.42. The z-score for a $500 margin would be (500-0)/141.42 = 3.54. Using a standard normal distribution table, we can find that the probability of a z-score less than 3.54 is approximately 0.9998. Therefore, the likelihood that the average income of the sample would be within $500 of the actual average income is approximately 0.9998 or 99.98%.

c) The difference between the answers to parts a) and b) is due to the larger sample size in part b). As the sample size increases, the standard error decreases, resulting in a narrower confidence interval around the actual average income. This means that the likelihood of the sample average being within a certain range of the population average increases as the sample size increases.

d) The suitable sample size would depend on the desired level of certainty and margin of error. If a higher level of certainty is desired, a larger sample size should be used. Similarly, if a narrower margin of error is required, a larger sample size is needed. The sample size also depends on the variability of the population. A higher variability may require a larger sample size to obtain a desired level of certainty.