Math 207 Solutions – Assignment 6

8.33

a. Since we can use a large-sample confidence interval for the population mean (since the distribution of is approximately normal). Because the confidence level is 99%, and (or you could also use 2.57 or 2.58). Then a 99% confidence interval for the mean weight of all packages sold in the smaller meat trays is: = (0.93 lbs., 1.09 lbs.).

b. We are 99% confident that the mean weight of all packages sold in the smaller meat trays is between 0.93 and 1.09 pounds. Our confidence is in the method we use: if the sampling were (hypothetically) done repeatedly, then 99% of the intervals created would contain the true mean weight. (Put another way, we used a method that gives us correct results 99% of the time.)

c. The 99% confidence interval contains the value 1. That is, 1 pound is among the likely values for the mean. Therefore, the quality control department has no reason to be concerned.

8.35

Since the population size of car registrations in California is obviously larger than 20 times the sample size of 500, we have approximate independence, and this is a binomial setting with n = 500 and p unknown. For our sample, .

a. Since and , we can use a large-sample confidence interval for p, the proportion of all California cars that are sports utility vehicles (since the distribution of is approximately normal). Because the confidence level is 95%, and (or you could also use 2, based on the Empirical Rule). Then a 95% confidence interval for p is: = (0.106, 0.166).

b. Estimating with a higher degree of accuracy means reducing the width of the confidence interval. This can be done by either using a lower confidence level or by taking a larger random sample. (Note there are tradeoffs for reducing the width of the interval: we either give up some confidence or we spend more time/money gathering more data.)

8.36

a. The completion time for the online order form probably does not have a mound-shaped distribution. It’s not possible to have a negative time, yet it is possible to have a very long completion time. Furthermore, most people probably won’t take very long to complete the form, but it’s possible that a few people (perhaps those computer-phobic folks) will take a very long time. Hence, I think the distribution is probably skewed right.

b. Even if the population of completion times is not normal, we know by the Central Limit Theorem (since n = 50 is “large”) that the distribution of the sample average times is approximately normal. With this information we can determine an approximate confidence interval for the population mean time.

c. Because the confidence level is 95%, and (or you could also use 2, based on the Empirical Rule). Then a 95% confidence interval for is: = (3.75 minutes, 5.25 minutes).

8.44

Note that both samples are of size 30, so they are “large” (). Then we can use the large-sample confidence interval for the difference in means and we can estimate the population standard deviations with the sample standard deviations. Since the confidence level is 95%, and (or 2, if you use the Empirical Rule).

Then a 95% confidence interval for is: = (6.75 pounds, 9.05 pounds).

We are 95% confident that the true difference in average weight losses (diet A – diet B) is between 6.75 and 9.05 pounds. Thus, we are 95% confident that people lose more weight on average with Diet A than with Diet B (since the interval is completely above 0). Our confidence is in the method we use: if the sampling were (hypothetically) done repeatedly and each time a confidence interval created, then 95% of the intervals would contain the true difference in average weight losses (i.e., we are using a method that gives us a correct answer 95% of the time).

Note: You could have equivalently solved this problem by differencing in the opposite way (diet B – diet A). Then the confidence interval would be (-9.05 pounds, -6.75 pounds) and the appropriate interpretation: We are 95% confident that the true difference in average weight losses (diet B – diet A) is between -9.05 and -6.75 pounds. Thus, we are 95% confident that people lose more weight on average with Diet A than with Diet B (since the interval is completely below 0).

8.58

Note that and , therefore is “large.” Also, and , therefore is “large.” Hence, we can use the large-sample confidence interval for a difference in population proportions. Since the confidence level is 95%, and (or 2, if you use the Empirical Rule).

Then a 95% confidence interval for is: = (-0.20, -0.02).

We are 95% confident that the true difference in proportions of union-backed winners (West – East) is between –0.20 and –0.02. Thus, we are 95% confident that the proportion of union-backed winners in the West is less than that of the East (since the interval is completely below 0). Our confidence is in the method we use: if the sampling were (hypothetically) done repeatedly and each time a confidence interval created, then 95% of the intervals would contain the true difference in proportions (i.e., we are using a method that gives us a correct answer 95% of the time). Note: Just as in problem 8.44, you could have done the differencing in the opposite direction, but if you interpret the interval correctly, then you reach the same conclusion.

8.74

This is a question about a difference in proportions (not about a single proportion). We don’t have any previous information with which we can estimate or in the margin of error. Hence, we can set them both equal to 0.5, which will give us a conservative estimate of n (i.e., the estimate of n may be larger than we need, but at least it won’t be too small—we are solving for n in the worst-case scenario). Also, we must assume both sample sizes are the same.

A confidence level is not stated in the question, but I asked you to use 95%. Then, and (or 2, if you use the Empirical Rule).We want the margin of error to be no more than 3 percentage points (i.e., no more than 0.03).

Hence, we need to find n such that

(always round up)

Therefore, both samples sizes should be at least 2135. (Note that this value of n is large enough, regardless of p, to justify our use of the large-sample confidence interval in the calculation.)

8.77

A confidence level is not stated in the question, but I asked you to use 99%. Then, and (or you could also use 2.57 or 2.58). We want the margin of error to be no more than 2 days.

Hence, we need to find n such that

(always round up)

Therefore, a sample of at least 166 hunters should be taken. (Note that this value of n is large enough to justify our use of the large-sample confidence interval in the calculation.)

8.81

This is a question about a difference in means (not about a single mean). The sample standard deviations of 24.3 micrograms per day and 17.6 micrograms per day can be used to estimate the population standard deviations. Since the confidence level is 90%, and (or you could also use 1.64 or 1.65). Furthermore, we must assume the samples are the same size and we want a margin of error of no more than 5.

We need to find n such that

(always round up)

Therefore, both sample sizes should be at least 98. (Note that this value of n is large enough to justify our use of the large-sample confidence interval in the calculation.)