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1、Solutions to Practice Problems for Part IV1. Five inspectors are employed to check the quality of components produced on an assembly line. For each inspector, the number of components that can be checked in a shift can be represented by a random variable with mean 120 and standard deviation 16. Let
2、X represent the number of components checked by an inspector in a shift. Then the total number checked is 5X, which has mean 600 and standard deviation 80. What is wrong with this argument? Assuming that inspectors performances are independent of one another, find the mean and standard deviation of
3、the total number of components checked in a shift.The calculation for the mean is OK, but the calculation of the standard deviation is not. Remember that a standard deviation is a square root, and that:We need to do our algebra on the variance, not the standard deviation. Therefore:2. It is estimate
4、d that in normal highway driving, the number of miles that can be covered by automobiles of a particular model on 1 gallon of gasoline can be represented by a random variable with mean 28 and standard deviation 2.4. Sixteen of these cars, each with 1 gallon of gasoline, are driven independently unde
5、r highway conditions. Find the mean and standard deviation of the average number of miles that will be achieved by these cars.3. A process is known to produce bricks whose weights are normally distributed with standard deviation 0.12 pounds. A random sample of sixteen bricks from todays output had a
6、 mean weight of 4.07 pounds.(a) Find a 99% confidence interval for the mean weight of all bricks produced today.A confidence interval consists of three elements: a measure of central tendency, a number of standard errors, and some measure of dispersion (e.g. a standard error). In this case the cente
7、r is 4.07 (our sample mean, which is the best estimate we have of central tendency). The number of standard deviations is 2.58 (found by looking in the z table where the probability is 0.495 - you might just as reasonably use 2.57 or 2.575). In this case we have a sample from a normal distribution w
8、ith a known standard deviation, so we can use Formula #1 (see the Confidence Interval Formulae sheet). The standard error is the basic standard error of the mean (): (b) Without doing the calculations, state whether a 95% confidence interval for the population mean would be wider than, narrower than
9、, or the same width as that found in (a).Narrower, because a larger alpha is associated with a narrower interval.(c) It is decided that tomorrow a sample of twenty bricks will be taken. Without doing the calculations, state whether a correctly calculated 99% confidence interval for the mean weight o
10、f tomorrows output will be wider than, narrower than, or the same width as that found in (a).Narrower, because a larger n makes the standard error smaller.(d) In fact, the population standard deviation from todays output is 0.15 pounds. Without doing the calculations, state whether a correctly calcu
11、lated 99% confidence interval for the mean weight of todays output will be wider than, narrower than, or the same width as that found in (a).Wider, because our estimated standard error was based on a standard deviation of 0.12. A larger standard error is associated with a wider interval.4. A product
12、ion manager knows that historically, the amounts of impurities in bags of a chemical follow a normal distribution with a standard deviation of 3.8 grams. A random sample of nine bags of the chemical yielded the following amounts of impurities in grams:18.213.715.917.421.816.612.318.816.2(a) Find a 9
13、0% confidence interval for the population mean weight of impurities.First, calculate . Again we have a sample from a normal distribution with a known standard deviation, so we can use Formula #1.(b) Without doing the calculations, state whether a 95% confidence interval for the population mean would
14、 be wider than, narrower than, or the same width as that found in (a).Wider, because a smaller alpha is associated with a wider interval.5. A random sample of 1,562 undergraduates enrolled in marketing courses was asked to respond on a scale from one (strongly disagree) to seven (strongly agree) to
15、the statement: Most advertising insults the intelligence of the average customer. The sample mean response was 3.92 and the sample standard deviation was 1.57.(a) Find a 90% confidence interval for the population mean response.In this case we have a large sample, so we can use Formula #1. The sample
16、 standard deviation s is used as a good estimate of the (unknown) population parameter m.(b) Without doing the calculations, state whether an 80% confidence interval for the population mean would be wider than, narrower than, or the same as (a).Narrower, because a larger alpha is associated with a n
17、arrower interval. (Remember that alpha is the area outside the confidence interval; as alpha gets bigger, the interval gets smaller.)6. The Cloze readability procedure is designed to measure the effectiveness of a written communication. (A score of 57% or more on the Cloze test demonstrates adequate
18、 understanding of the written material.) A random sample of 352 certified public accountants was asked to read financial report messages. The sample mean Cloze score was 60.41% and the sample standard deviation was 11.28%. Find a 90% confidence interval for the population mean score, and comment on
19、your result.As was the case in the previous problem, the large sample size allows us to use Formula #1.We can safely assume that the financial report messages are effectively written.7. A population has a normal distribution with unknown mean and unknown variance. We know how to find a confidence in
20、terval for the population mean, given a random sample of two observations. We do not, however, know how to find such a confidence interval with a random sample of only one. Why not?We have no way to estimate the variance. Not only does the denominator of the variance formula call for (n - 1), which,
21、 in this case would require us to divide by zero, but the t statistic is undefined with zero degrees of freedom.8. In October 1992, ownership of the San Francisco Giants baseball team considered a sale of the franchise that would lead to a move to Florida. A random sample of 610 San Francisco Bay Ar
22、ea taxpayers, carried out by the San Francisco Examiner, contained 50.7% who would be disappointed by this move. Find a 99% confidence interval for the population proportion of Bay Area taxpayers with this feeling.This is a proportion problem with a large sample; we will use Formula #3:9. A random s
23、ample was taken of 189 National Basketball Association games in which the score was not tied after one quarter. In 132 of these games, the team leading after one quarter won the game. (a) Find a 90% confidence interval for the population proportion of all occasions on which the team leading after on
24、e quarter wins the game.Another proportion problem with a large sample. Note that .Of all the games that are not tied after one quarter, we are 90% sure that the proportion of games that are eventually won by the team leading after the first quarter is somewhere between 64.35% and 75.33%.(b) Without
25、 doing the calculations, state whether a 95% confidence interval for the population proportion would be wider than or narrower than that found in (a).Wider, because a smaller alpha is associated with a wider interval.10. Of a random sample of 323 union members, 47.9% agreed or strongly agreed with t
26、he statement: Union workers should refuse to work when a nonunion worker is sent to the job. Based on this information, a statistician calculated, for the percentage of all union members with this view, a confidence interval running from 45.8% to 50.0%. Find the level of confidence associated with t
27、his interval.This is like the previous two problems, but we are working backwards. Instead of being given a desired confidence level and having to find the upper and lower limits of the interval, we are doing the reverse. Note that the upper and lower boundaries of the confidence interval are 2.1% (
28、or 0.021) away from the estimated proportion of .479. This means that:0.021And therefore The confidence level of this interval is 55.28%.11. A random sample was taken of 96 foreign manufacturers, with direct investment in the United States, who use independent U.S. industrial distributors. Of these
29、sample members, 32 said the distributors were rarely or never capable of performing the advice and technical support function. Find an 80% confidence interval for the population proportion.Another problem for Formula #3. Note that .Or 12. For a random sample of 40 accounting students in a class usin
30、g group learning techniques, the mean examination score was 322.12, and the sample standard deviation was 54.53. For an independent random sample of 61 students in the same course but in a class not using group learning techniques, the sample mean and standard deviation of the scores were 304.61 and
31、 62.61, respectively. Find a 95% confidence interval for the difference between the two population mean scores.We are studying the difference between two sample means, which means we will use either Formula #4, or #5. We cant use the matched pairs formula because the two samples are different sizes,
32、 so we use #5:Or (-5.56, 40.58)13. In a survey of practicing certified public accountants on women in the accounting profession, sample members were asked to respond on a scale of one (strongly disagree) to five (strongly agree) to the statement: Women are equally acceptable to clients as are men to
33、 perform work on engagements. For a random sample of 172 female accountants, the mean response was 3.483 and the sample standard deviation was 0.970. For an independent random sample of 186 male accountants, the sample mean and standard deviation were 3.435 and 0.894, respectively. Find a 95% confid
34、ence interval for the difference between the two population means.Same as the previous problem; use Formula #5.Or (-0.1457, 0.2417)14. For a random sample of 190 firms that revalued their fixed assets, the mean ratio of debt to tangible assets was 0.517 and the sample standard deviation was 0.148. F
35、or an independent random sample of 417 firms that did not revalue their fixed assets, the mean ratio of debt to tangible assets was 0.489 and the sample standard deviation was 0.159. Find a 99% confidence interval for the difference between the two population means.Formula #5:Or (-0.0062, 0.0622)15.
36、 Of a random sample of 1203 business students in 1979, 20.2% said that teaching, as a career was very unappealing. Of an independent random sample of 1203 business students taken in 1989, 13.2% had this reaction to teaching as a career. Find a 99% confidence interval for the difference between the p
37、opulation proportions regarding teaching as very unappealing in the two years.Now we have the difference between two proportions, so we use Formula #6:Or (0.031, 0.109)16. Supermarket shoppers were observed, and questioned immediately after putting an item in their cart. Of a random sample of 570 ch
38、oosing a product at the regular price, 308 claimed to check price at the point of choice. Of an independent random sample of 232 choosing a product at a special price, 157 made this claim. Find a 90% confidence interval for the difference between the two population proportions.Formula #6:Or (-0.198,
39、 -0.076)17. Referring back to Problem 3 above, find the sample size needed so that a 90% confidence interval for the population mean weight of bricks extends an amount 0.01 on each side of the sample mean.Therefore the sample must be18. A politician wants to estimate the proportion of constituents f
40、avoring a controversial piece of proposed legislation. Suppose that a 90% confidence interval that extends at most 0.05 on each side of the sample proportion is required. How many sample observations are needed?Using our formula from the notes for Part IV:Therefore the sample must be19. An investiga
41、tor wants to compare two population proportions and intends to take independent random samples of the same size from each population. She wants to be sure that a 90% confidence interval for the difference between the two population proportions extends no further than 0.05 on each side of the differe
42、nce between the sample proportions. How large a sample should she take from each of the two proportions?This is tricky, because we dont have a formula in the notes to answer this exact problem. (We have sample size formulas for a single mean and a single proportion, but not for the difference betwee
43、n two means, or, as is the case here, the difference between two proportions.) We will adapt what we know to address this situation.Recall that the variance of the sum of two random variables is the sum of their variances. Similarly, the variance of a difference between two random variables is the s
44、um of their variances. Therefore the variance of the difference between two sufficiently large sample proportions is:Now, using our formula from the notes for Part IV:nTherefore, she will need 542 from each population. Note that we never had to get an estimate for p to do this calculation. We know t
45、hat the product of p(1 - p) can never exceed 0.25, so we plug that value in and get an upper bound on our sample size.20. In a random sample of 1,158 newly promoted executives, 47.9% rated a statistics course as very important or somewhat important as part of the preparation for a career in general
46、management. (a) Find a 99% confidence interval for the population proportion of all newly promoted executives holding this view.Using Formula #3,Or (b) Based on this sample information, a statistician computed a confidence interval for the population proportion running from 0.458 to 0.500. What is t
47、he probability content of this interval?This is similar to the problem above regarding 323 union members, in that we will be working backwards to arrive at a confidence level. Each of these limits is 0.021 away from 0.479.21. Of a random sample of 151 marketing executives in consumer goods manufactu
48、ring, 76.0% said that brand identification held by incumbents was an important or extremely important barrier to entering new markets. Based on this information, a statistician computed, for the population proportion with this view, the confidence interval Find the probability content of this interval.Same as the previous problem. Note that the center of this interval
