# Statistics 578 Assignemnt 3

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Assignment-3-solution: (Chs. 7 and 8):Due by Midnight of Sunday, October 14th, 2012: drop box 3: Total 70 points True/False (1 point each) Chapter 7 1. A sample size of 1000 is large enough to conclude that the sampling distribution of p is a normal distribution, when the estimate of the population proportion is .996. FALSE Here, n(1-p) = 1000*0.004 = 4 which is less than 10 or even 5 (the liberal rule suggested in the book) 2. The standard deviation of all possible sample proportions decreases as the sample size decreases. FALSE It increases when n gets smaller. 3. If the population is normally distributed then the sample mean is normally distributed for any sample size. TRUE (Instructions on Ch 7, property 4) 4. The reason sample variance has a divisor of n-1 rather than n is that it makes the standard deviation an unbiased estimator of the population standard deviation. FALSE Sample variance is unbiased but sample standard deviation is not.

5. The mean of the sampling distribution ofsampled population. TRUE

is always equal to the mean of the

Chapter 86. First a confidence interval is constructed without using the finite population correction factor. Then, for the same identical data, a confidence interval is constructed using the finite population correction factor. The width of the interval with the finite population correction factor is wider than the confidence interval without the finite population correction factor. False. Look at the formula for finite population correction. It reduces the standard deviation.

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7. When the population is normally distributed and the population standard deviation is unknown, then for any sample size n, the sampling distribution of is based on the t distribution. TRUE When we substitute the estimated standard deviation in the formula, the distribution becomes t-distribution instead of Z distribution. 8. When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on a sample of n=200 will be wider than a confidence interval for a population mean based on a sample of n= 150. False 9. When the level of confidence and the sample size remain the same, a confidence interval for a population mean will be narrower, when the sample standard deviation s is small than when s is large. True 10. When the level of confidence and sample proportion p remain the same, a confidence interval for a population proportion p based on a sample of n=100 will be narrower than a confidence interval for p based on a sample of n=400. FALSE 11. The sample mean, the sample proportion and the sample standard deviation are all unbiased estimators of the corresponding population parameters. FALSE The Sample Standard Deviation is not an unbiased estimator (Clearly stated in Instructions on Ch8 page6) 12. Assuming the same level of significance , as the sample size increases, the value of t/2 approaches the value of z/2. TRUE Multiple Choices (2 points each) Chapter 7 1. A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population mean and standard deviation of 90 lbs. and 24 lbs. respectively, then based on a sample size of 36 boxes, the probability that the average weight of the boxes will exceed 94 lbs. is: A. 15.87% B. 84.13% C. 34.13%2

D. 56.36% E. 16.87%The

= 24/36 = 4. Therefore the Z score is (94-90)/4 = 1.0

P(Z 1.0) = 1 - 0.8413 = 0.1587 or 15.87%.

2. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years 13 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What is the probability that more than 40 accounts will be classified as delinquent? A. 42.07% B. 92.07% C. 7.93% D. 40.15% E. 90.15% Here = 0.13, n = 250, and 1- = 0.87. We have n = 32.5 and n(1- ) = 217.5, both greater than 10, and n (1- ) = 28.275 > 10. So, normal approximation without continuity correction is appropriate. The standard error of p = p = = 0.0213

Next, 40 accounts out of 250 in proportion is 40/250 = 0.16. The question refers to more than 40. Therefore, the question is P(p 0.16)? Using the standardization process with p = 0.13 (the population proportion) and p = 0.0213, we have P(Z ) = P(Z 1.41) = 1- P(Z 1.41) = 10.9207 = 0.0793 from the table or MegaStat. 3. If we have a sample size of 100 and the estimate of the population proportion is 0.10, the mean of the sampling distribution of the sample proportion is: A. 0.009 B. 0.10 C. 0.03

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D. 0.90 E. 0.09 4. Consider a sampling distribution formed based on n = 10. The standard deviation of the population of all sample means is ______________ more than the standard deviation of the population of individual measurements: . A. Always B. Sometimes C. Never Should be obvious from the formula for the standard deviation

5. A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population mean and standard deviation of 90 lbs. and 24 lbs. respectively, then based on a sample size of 36 boxes, the probability that the average weight of the boxes will be be less than 84 lbs. is: A. 16.87% B. 93.32% C. 43.32% D. 6.68% E. 84.13% The = 24/36 = 4. Therefore the Z score is (84-90)/4 = -1.5 P(Z -1.5) = 0.0.0668 6. If a population distribution is known to be normal, then it follows that: A. The sample mean must equal the population mean B. The sample mean must equal the population mean for large samples C. The sample standard deviation must equal the population standard deviation D. All of the above E. None of the above 7. According to a hospital administrator, historical records over the past 10 years have shown that 20% of the major surgery patients are dissatisfied with aftersurgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted.4

Sixty-four (64) patients indicated that they were dissatisfied with the after surgery care. What are the mean and the standard deviation of the sampling distribution of p ? A. 16% and .034% B. 20% and 1.83% C. 20% and 2% D. 20% and .034% E. 20% and 16% Chapter 8 8. The width of a confidence interval will be: A. Narrower for 95% confidence than 99% confidence B. Wider for a sample size of 100 than for a sample size of 50 C. Wider for 90% confidence than 95% confidence D. Wider when the sample standard deviation (s) is small than when s is large 9. When the level of confidence and sample size remain the same, a confidence interval for a population proportion p will be __________ when p(1-p) is larger than when p(1-p) is smaller. A. Narrower B. Wider C. Neither A nor B, they will be the same D. Cannot tell from the information given 10. A confidence interval increases in width as A. The level of confidence increases B. n decreases C. s increases D. All of the above E. None of the above 11. In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a standard deviation of .3 inches. What is the 95% confidence interval for the true mean length of the bolt?5

A. 2.804 to 3.196 B. 2.308 to 3.692 C. 2.769 to 3.231 D. 2.412 to 3.588 E. 2.814 to 3.186 Since the population standard deviation is not given (only sample std. dev given) and the sample is small, t distribution is appropriate. Using MegaStat (you could use t Table too) Confidence interval - mean 95% confidence level 3 mean 0.3 std. dev. 9 n 2.306 t (df = 8) 0.231 half-width upper confidence 3.231 limit lower confidence 2.769 limit

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12. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent (more than 90 days overdue). For this quarter, the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent. What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company? A. .1485 to .2515 B. .1608 to .2392 C. .1992 to .2008 D. .1671 to .2329 E. .1714 to .2286

Confidence interval :proportion 95% confidence level 0.2 proportion 400 n 1.960 z 0.039 half-width 0.239 upper confidence limit 0.161 lower confidence limit

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13. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are current (between 0 and 60 days after billing). The historical records show that over the past 8 years 70 percent of the accounts have been current. Determine the sample size needed in order to be 95% confident that the sample proportion of the current customer accounts is within .02 of the true proportion of all current accounts for this company. A. 1421 B. 1549 C. 1842 D. 2017 E. 3484

You can use the formula given in my instructions or book or MegaStat. We always round up the required sample size to the whole number because the formula gives the minimumSample size - proportion0.02 0.7 95% 1.960 2016.766 2017 E, error tolerance estimated population proportion confidence level z sample size rounded up

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14. A company is interested in estimating the mean number of days of sick leave taken by its employees. The firm's statistician randomly selects 200 personnel files and notes the number of sick days taken by each employee. The estimated sample mean is 12.2 days and the sample standard deviation is 10 days. How many personnel files would the director have to select in order to estimate to within 1 day with a 99% confidence interval? A. 166 A. 200 B. 385 C. 550 E. 664

Sample size - mean1 10 99% 2.576 663.490 664 E, error tolerance standard deviation confidence level z sample size rounded up

Essay Type (show your work) (5 points each) Chapter 7 1. A PGA (Professional Golf Association) tournament organizer is attempting to determine whether hole (pin) placement has a significant impact on the average number of strokes for the 13th hole on a given golf course. Historically, the pin has been placed in the front right corner of the green and the historical mean number of strokes for the hole has been 4.25, with a standard deviation of 1.6 strokes. On a particular day during the most recent golf tournament, the organizer placed the hole (pin) in the back left corner of the green. 49 golfers played the hole with the new placement on that day. Determine the probability of the sample average number of strokes more than 4.55.9

Here = 4.25, = 1.6, and n= 49. So,

= 1.6/49 = 0.2286. Therefore,

Z = (4.55 - 4.25)/0.2286 = 1.31 (rounded) The right hand side area or probability is required because the question is ..more than 4.55. So we have to subtract from 1 the probability of the calculated Z to the left (given in the table or by MegaStat).P( 4.65) = 1- P(Z 1.31) = 1- 0.9049 (using table or computer) = 0.0951

2. The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.27 inches. A sample of 9 metal sheets is randomly selected from a batch. What is the probability that the average length of a sheet is between 30.23 and 30.32 inches long? The Standard deviation of the sample mean = 0.27/9 = 0.09. Therefore, P{(30.23 30.05)/0.09 Z (30.32 30.05)/0.09} = P(2 Z 3) = 0.99865 -0.9772 = 0.02145 3. Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of 0.18 ounces. The weights of the sugar bags are normally distributed. What is the probability that 36 randomly selected packages will have average weight less than 15.955 ounces? P[Z (15.955 16)/(0.18/36) ] = P(Z -1.5) = 0 .0668. Here left side probability or area is required (less than). So, we do not subtract from 1. Chapter 8 4. A small town has a population of 20,000 people. Among these 1,000 regularly visit a popular local bar. A sample of 100 people who visit the bar is surveyed for their annual expenditures in the bar. It is found that on average each person who regularly visits the bar spends about $2000 per year in the bar with a standard deviation of $200. Construct a 99 percent confidence interval around the mean annual expenditure in the bar.

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This problem needs finite population correction because the population from which sample is taken is only 1000 not 20,000 (that number was given to trick you) and is Not at least 20 times the sample of 100. The finite population correction factor (given in my Instructions pages 5 and 6; the book gives a wrong formula, perhaps a type mistake) is (N-n)/(N-1) or 900/999 in this case or 0.901. Its square root is 0.949. So,

=

= (200/10)*0.949 = 18.98.

For 99% confidence, the margin using normal distribution (because df is quite large) is: 2.576*18.98 = 48.89. Therefore, the interval is 2000 48.89 or between $1951.11 and $2048.89 or $1951 and $2049 rounded. If you use the theoretically more correct t-distribution (because the population std dev is not given), then t for 99% confidence interval (or t.005) would be 2.626 which would give the interval between $1950 and $2050, a difference of (less than) a dollar in amounts around two thousand dollars. Therefore, I accept both results, although t-distribution is theoretically more correct when population std dev is not known, as discussed in my Instructions. 5. The production manager for the XYZ manufacturing company is concerned that the customer orders are being shipped late. He asked one of his planners to check the timeliness of shipments for 1000 orders. The planner randomly selected 1000 orders and found that 100 orders were shipped late. Construct the 99% confidence interval for the proportion of orders shipped late. Here, p = 100/1000 = 0.10. Also, np(1-p) = 90 > 10. Therefore, normal approximation without continuity correction can be done. The confidence interval for the proportions of orders shipped late using Z value of 2.576 for 99% confidence is: (100/1000) 2.576 [(0.10*0.90)/1000] = 0.076 to 0.124 (rounded) Or using MegaStat:

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Confidence interval - proportion99% 0.1 1000 2.576 0.024 0.124 0.076 confidence level proportion n z half-width upper confidence limit lower confidence limit

6. The weight of a product is measured in pounds. A sample of 49 units is taken from a batch. The sample yielded the following results: = 80 lbs. and s=10.5 lbs. Calculate a 90 percent confidence interval for the mean. The t-value for 90% confidence interval with df 48 is 1.677. Therefore, the confidence interval is 80 1.677(10.5/49) = 77.48 to 82.52 (rounded). However, if a student followed the practical rule and calculated using Z value as follows: 80 1.645(10/50) = 77.53 to 82.47. I accepted it. You can see that the difference is in second decimal which is negligible considering the large integral (whole number) values.

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