QA 233 (Basic Business Statistics)

Solutions to Practice Problem Set VIII

1. When surveyed, 41 out of 200 respondents indicate that ‘Twinkies’ are their favorite snack food. Provide both a point estimate and an interval estimate (at the 92% level of confidence) of the share of the population for whom ‘Twinkies’ are their favorite snack food. At the 90% level of confidence, what is the margin of error?

   Our random variable is whether or not a consumer’s favorite snack food is ‘Twinkies,’ which is a NOMINAL level variable. Thus we should consider using the proportion to summarize this data. We do not ‘know’ the population, but instead are given the results of a sample. The sample proportion who indicated that ‘Twinkies’ are their favorite snack food is

   

   The first part of the question asks us to provide both a point estimate and an interval estimate (at the 92% level of confidence) of the share of the population for whom ‘Twinkies’ are their favorite snack food. The point estimate of the population proportion is the sample proportion p, which we have already calculated. Keep in mind that we do not ‘know’ the population, but are trying to estimate a population parameter - this means that the interval estimate of the share (proportion) of the population for whom ‘Twinkies’ are their favorite snack food is a confidence interval. Since we don't 'know' the population, we cannot really say if the sample is sufficiently large to ensure (by the Central Limit Theorem) that the sampling distribution of p is normal - but we can use p = 0.50 (the most conservative value) as a check. Using p = 0.50 we have np = 200(0.50) ³ 5, n(1-p) = 200(1 - 0.50) ³ 5) so the sample is certainly sufficiently large  to ensure (by the Central Limit Theorem) that the sampling distribution of p is normal.  The 92% confidence interval estimate of a population proportion p is given by

   

   which yields an interval of

   

   for this problem. The second part of the question asks us to provide, at the 90% level of confidence, the margin of error. The margin of error (or half-width of a confidence interval) is given by:

   

   which yields a value of

   

   for this problem.

2. Suppose that the mean sales for nineteen randomly selected restaurants in the ‘Hamburger Haven’ chain (which consists of 800 restaurants) during the week of the Thanksgiving holiday are $9,250.00. If the standard deviation of these sample results is $1250.00, what is the 95% confidence interval?

   Our random variable is sales for a restaurant in the ‘Hamburger Haven’ chain during the week of the Thanksgiving holiday, which is a RATIO level variable. Thus we should consider using the mean to summarize this data. We do not ‘know’ the population, but instead are given the results of a sample. The sample mean and standard deviation that we are given (based on the sample of n = 19 restaurants) are:

   x = 9250.00 and s = 1250.00

   This question asks us to provide an interval estimate (at the 95% level of confidence) of the mean sales for all restaurants in the ‘Hamburger Haven’ chain (our universe of 800 restaurants) during the week of the Thanksgiving holiday. Again, keep in mind that we do not ‘know’ the population, but are trying to estimate a population parameter - this means that the interval estimate of the mean sales of the population is a confidence interval. Now we have a small sample but we do not know if the population (sales for restaurants in the ‘Hamburger Haven’ chain during the week of the Thanksgiving holiday) is normally distributed. If the population (sales for restaurants in the ‘Hamburger Haven’ chain during the week of the Thanksgiving holiday) were known to be normally distributed, we could build a confidence interval estimate of the population mean m by using:

   

   with n - 1 = 18 degrees of freedom for t. Or, if the sample was sufficiently large (n ³ 30), we could build a confidence interval estimate of the population mean m by using:

   

   However, the necessary conditions are not met for either of these approaches. We do not have any means (that we have discussed in class) by which we can construct a confidence interval estimate of the population mean m under the conditions given in this problem.

3. Suppose that the mean sales for nineteen randomly selected restaurants in the ‘Hamburger Haven’ chain (which consists of 800 restaurants) during the week of the Thanksgiving holiday are $9,250.00. If the standard deviation of these sample results is $1250.00 and mean weekly sales for restaurants in the ‘Hamburger Haven’ chain tends to be normally distributed, what is the 95% confidence interval?

   Our random variable is sales for a restaurant in the ‘Hamburger Haven’ chain during the week of the Thanksgiving holiday, which is a RATIO level variable. Thus we should consider using the mean to summarize this data. We do not ‘know’ the population, but instead are given the results of a sample. The sample mean and standard deviation that we are given (based on the sample of n=19 restaurants) are:

   x = 9250.00 and s = 1250.00

   This question asks us to provide an interval estimate (at the 95% level of confidence) of the mean sales for all restaurants in the ‘Hamburger Haven’ chain (our universe of 800 restaurants) during the week of the Thanksgiving holiday. Again, keep in mind that we do not ‘know’ the population, but are trying to estimate a population parameter - this means that the interval estimate of the mean sales of the population is a confidence interval. However, we have a small sample and we now do know that the population is normally distributed. Thus we can build a confidence interval estimate of the population mean m by using:

   

   with n - 1 = 18 degrees of freedom for t, which yields an interval of

   

4. If the actual mean weekly sales restaurants in the ‘Hamburger Haven’ chain is $11,200.50 and the actual standard deviation of weekly sales for restaurants in the ‘Hamburger Haven’ chain is $2,898.43, what is the probability that a randomly selected sample of 43 restaurants in the ‘Hamburger Haven’ will yield a mean of $12,500.00 or greater? Under these conditions, what is the probability that a randomly selected sample of 43 restaurants in the ‘Hamburger Haven’ will yield a mean between $12,000.00 and $12,230.00?

   Our random variable is again sales for a restaurant in the ‘Hamburger Haven’ chain, which is a RATIO level variable. Thus we should consider using the mean to summarize this data. However, in this problem we do ‘know’ the population, i.e., we are given that

   m = 11200.50 and s = 2889.43

   which implies that we are not dealing with confidence intervals. The first part of the problem asks us to find the probability that a randomly selected sample of 43 restaurants in the ‘Hamburger Haven’ chain will yield a mean of $12,500.00 or greater. The sample is sufficiently large (n = 43 ³ 30) to ensure (by the Central Limit Theorem) that the sampling distribution of x is normal. Thus the question can be rewritten as

   p(x ³ 12500) = ?

   By using the standard normal (z) transformation, we can answer this question. The transformation and solution are:

   

   The second part of the problem asks us to find the probability that a randomly selected sample of 43 restaurants in the ‘Hamburger Haven’ chain will yield a mean between $12,000.00 and $12,230.00. The sample is sufficiently large (n = 43 ³ 30) to ensure (by the Central Limit Theorem) that the sampling distribution of x is normal. Thus the question can be rewritten as

   p(12000 £ x £ 12230) = ?

   By using the standard normal (z) transformation, we can answer this question. The transformation and solution are:

   

5. If 5% of all output produced by a machine is defective, what is the probability that the proportion of defectives in the output produced by the machine today (300 units) is between .04 and .07? If our policy is perform maintenance on the machine when the proportion of defects in one day’s production exceeds 0.06, what is the probability that we will have to call maintenance at the end of any day? Given your results, what do you think of their maintenance policy (to perform maintenance on the machine when the proportion of defects in one day’s production exceeds 0.06)?

   Our random variable is whether or not a unit produced by this machine is defective, which is a NOMINAL level variable. Thus we should consider using the proportion to summarize this data. In this case we do ‘know’ the population, i.e., we are given that

   p = 0.05

   which implies that we are not dealing with confidence intervals. In the first part of the problem we are asked to find the probability that the proportion of defectives in the output produced by the machine today (300 units) is between 0.04 and 0.07. The sample is sufficiently large (np = 300(0.05) ³ 5, n(1-p) = 300(1 - 0.05) ³ 5) to ensure (by the Central Limit Theorem) that the sampling distribution of p is normal. Thus the question can be rewritten as

   p(0.04 £ p £ 0.07) = ?

   By using the standard normal (z) transformation, we can answer this question. The transformation and solution are:

   

   In the second part of the problem we are asked to find the probability that the proportion of defectives in the output produced by the machine today (300 units) exceeds 0.06. Again, the question can be rewritten as

   p(p ³ 0.06) = ?

   By using the standard normal (z) transformation, we can answer this question. The transformation and solution are:

   

   Under these conditions, the policy will most likely result in many maintenance calls.

6. Suppose that the mean checking account balance at a local bank is $750.00, and the standard deviation of checking account balances at the local bank is $500.00. Construct an interval that will contain 90% of all possible sample mean checking account balances that can result from a random sample of 45 checking accounts taken from the local bank.

   Our random variable is the checking account balance for individual accounts at the local bank, which is a RATIO level  (or INTERVAL level if we allow for overdrawn accounts) variable. Thus we should consider using the mean to summarize this data. However, in this problem we do ‘know’ the population, i.e., we are given that

   m = 750.00 and s = 500.00

which implies that we are not dealing with confidence intervals. We are asked to find the interval that will contain 90% of all possible sample mean checking account balances that can result from a random sample of 45 checking accounts at the local bank. The sample is sufficiently large (n = 45 ³ 30) to ensure (by the Central Limit Theorem) that the sampling distribution of x is normal. Thus the question can be rewritten as

By substitution we have:

So the interval is (627.76, 872.24).

7. Farmer Douglass, who has an orchard of 2,000 pomegranate trees, wants to estimate the proportion of trees whose fruit are ripe. He and his able bodied assistant Eb randomly select 35 trees and determine that the fruit of 19 of the sampled trees is ripe. Farmer Douglass now wants an interval estimate, at the 99% confidence level, of the proportion of trees in his orchard whose fruit are ripe. Please provide farmer Douglass with this estimate. Do you think that this interval estimate will be useful to farmer Douglass? Why or why not?

Our random variable is whether or not a tree in farmer Douglass’ orchard has ripe fruit, which is a NOMINAL level variable. Thus we should consider using the proportion to summarize this data. We do not ‘know’ the population, but instead are given the results of a sample. The sample proportion of trees in farmer Douglass’ orchard that have ripe fruit is

The question asks us to provide an interval estimate (at the 99% level of confidence) of the proportion of trees in farmer Douglass’ orchard that have ripe fruit. Keep in mind that we do not ‘know’ the population, but are trying to estimate a population parameter - this means that the interval estimate of the proportion of the population (farmer Douglass’ orchard) that has ripe fruit is a confidence interval. The 98% confidence interval estimate of a population proportion p is given by

which yields an interval of

for this problem.

This interval is quite wide and will probably not be of much use to farmer Douglass. To generate a narrower interval, farmer Douglass must either i) send Eb out to take a larger sample or ii) use a smaller level of confidence.

8. Suppose that 82% of all investors in the bond market earned a profit during the past fiscal year. Local bond dealer E. Z. Cash has seventy investors in the bond market, fifty of whom earned a profit during the past fiscal year. If we randomly select seventy investors, what is the probability the proportion of these investors who earned a profit during the past fiscal year would be at least as low as the proportion of E. Z.’s investors who earned a profit during the past fiscal year?

Our random variable is whether or not an investor in the bond market earned a profit during the past fiscal year, which is a NOMINAL level variable. Thus we should consider using the proportion to summarize this data. In this case we do ‘know’ the population, i.e., we are given that:

p = 0.82

which implies that we are not dealing with confidence intervals. In this problem we are asked to find the probability that the proportion of seventy randomly selected investors who earned a profit during the past fiscal year would be at least as low as the proportion of E. Z.’s investors who earned a profit during the past fiscal year? The proportion of E. Z.’s investors who earned a profit during the past fiscal year is:

The sample is sufficiently large (np = 70(0.82) ³ 5, n(1 - p) = 70(1 - 0.82) ³ 5) to ensure (by the Central Limit Theorem) that the sampling distribution of p is normal. Thus the question can be rewritten as:

p(p £ 0.714) = ?

By using the standard normal (z) transformation, we can answer this question. The transformation and solution are:

Thus it is unlikely that the investors would have done as poorly using randomly selected brokers as they did using E. Z.

9. A computer equipment manufacturer advertises that their 2500-L laser jet printer averages 12 printed pages per minute with a standard deviation of 1.50 pages per minute. You randomly select 50 documents from your hard drive and measure the number of pages printed in the first minute of printing for each document. The mean number of pages printed in the first minute for these 50 documents is 11.4. If the advertisement is accurate, what is the probability of getting results at least as poor as those you obtained (i.e., that the print speed is no more than 11.4 pages per minute for the first minute of printing for 50 randomly selected documents)?

Our random variable is the number of pages printed during the first minute of printing a document, which is a RATIO level variable. Thus we should consider using the mean to summarize this data. We do ‘know’ the population (or at least what the manufacturer tells us), i.e., we are given that:

m = 12.0 and s = 1.50

which implies that we are not dealing with confidence intervals. We are asked to find the probability of getting results at least as poor as those you obtained (i.e., that the print speed is no more than 11.4 pages per minute for the first minute of printing for 50 randomly selected documents). The sample is sufficiently large (n = 50 ³ 30) to ensure (by the Central Limit Theorem) that the sampling distribution of x is normal. Thus the question can be rewritten as:

p(x £ 11.4) = ?

By using the standard normal (z) transformation, we can answer this question. The transformation and solution are:

The results suggest that the manufacturer’s claim is not true.

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