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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
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