MGT 218 : Study Guide for chapter 5 , 6 , 7 , 8
MGT 218 : Study Guide for chapter 5 , 6 , 7 , 8 MGT 218
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Study Guides Mid-term Exam MGT 218 : Chapter 5 , 6 ,7 ,8 and solution for exercises on the text book. Book (Introductory Statistics 8e edition : Prem S.Mann) Chapter 5 : Discrete Random Variables and their Probability Distributions Exercise ( Even number exercises ) Explaining Important Most Important Theory: Random variable is a variable whose value is determined by the outcome of random experiment. Ex : the number of fish caught on a fishing trip. Discrete Random variable : that assumes countable values Ex : the number of heads obtained in three tosses of a coin Continuous Random variable : a random variable that can assume any value contained in one or more intervals. Probability Distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities 5.24) Find the mean and standard deviation for each of the following probability distributions: \ a) x P(x) x. P(x) x*x x*x*P (x) 3 .09 .27 9 .81 4 .21 .84 16 3.36 5 .34 1.7 25 8.5 6 .23 1.38 36 8.28 7 .13 .91 49 6.37 Mean = 5.1 Standard deviation = 1.14 b) x P(x) x . P(x) x*x x*x* P(x) 0 .43 0 0 0 1 .31 .31 1 .31 2 .17 .34 4 .68 3 .09 .27 9 .81 Mean = 0.92 Standard deviation = 0.98 5.26 ) Let x be the number of magazines a person reads every week. Based on a sample survey of adults, the following probability distribution table was prepared.Find the mean and the standard deviation of x x P(x) x . P(x) x*x x*x* P(x) 0 .36 0 0 0 1 .24 .24 1 .24 2 .18 .36 4 .72 3 .1 .3 9 .9 4 .07 .28 16 1.12 5 .05 .25 25 1.25 Mean = 1.43 Standard deviation = 1.48 5.28 ) The following table gives the probability distribution of the number of camcorders sold on a given day at an electronics store : Patients per x . P hour Probability (x) x * x x *x * P(x) 0 0.2725 0 0 0 0.354 1 0.3543 3 1 0.3543 0.460 2 0.2303 6 4 0.9212 0.299 3 0.0998 4 9 0.8982 0.129 4 0.0324 6 16 0.5184 5 0.0084 0.042 25 0.21 0.013 6 0.0023 8 36 0.0828 Mean = 1.3 Standard deviation = 1.1 5.30 ) Let x be the number of potential weapons detected by a metal detector at an airport on a given day. The following table lists the probability of distribution of x. x . P X P (x) (x) x * x x *x * P(x) 0 0.14 0 0 0 1 0.28 0.28 1 0.28 2 0.22 0.44 4 0.88 3 0.18 0.54 9 1.62 4 0.12 0.48 16 1.92 5 0.06 0.3 25 1.5 Mean = 2.04 Standard deviation = 1.43 5.32 ) Refer to Exercise 5.15. Find the mean and standard deviation of the probability distribution you developed for the number of remote starting systems installed per day by Al’s Auto Securtity Shop over the past 80 days. Give a brief of the values of mean and standard deviation. x . P x P(x) (x) x * x x *x * P(x) 1 0.1 0.1 1 0.1 2 0.25 0.5 4 1 3 0.3 0.9 9 2.7 4 0.2 0.8 16 3.2 5 0.15 0.75 25 3.75 Mean = 3.05 Standard deviation = 1.21 5.34) Refer to the probability distribution you developed in Exercise 5.17 for the number of uninsured motorists in a sample of two motorists. Calculate the mean and standard deviation of x for that probability distribution. x P(x) x . P (x) x * x x *x * P(x) 0 0.703921 0 0 0 1 0.135079 0.135079 1 0.135079 2 0.025921 0.051842 4 0.103684 Mean = 0.19 Standard deviation = 0.45 5.36 )** 8 An instant lottery ticket costs $2. Out of a total of 10,000 tickets printed for this lottery, 1000 tickets contain a prize of $5 each, 100 tickets have a prize of $10 each, 5 tickets have a prize of $1000 each, and 1 ticket has a prize of $5000. Let x be random variable that denotes the net amount a player wins by playing this lottery. Write the distribution of x. Determine the mean and standard deviation of x. How will you interpret the values of the mean and standard deviation of x. x P(x) x . P (x) x * x x *x * P(x) 5-2 1000/10000 0.3 9 0.9 10-2 100/10000 0.08 64 0.64 1000-2 5/10000 0.499 996004 498.002 5000-2 1/10000 0.4998 24980004 2498.0004 Mean = 1.38 Standard deviation = 54.73 Chapter 6 : Continuous Random Variables and the Normal Distribution ( Part 1 ) with the Even number exercises. The probability that a continuous random variable x assumes a single value is always zero. or a normal curve is a bell-shaped ( symmetric ) curve. A normal probability distribution, when plotted, gives a bell-shaped curve such that : a) The total area under the curve is 1.0 b) The curve is symmetric about the mean c) The two tails of the curve extend indefinitely. Standard normal distribution : is a special case of the normal distribution. With the normal distribution with mean = 0 and standard normal distribution = 1 z Values or z Scores : the units marked on the horizontal axis of the standard normal curve are denoted by z and are called the z values or z scores. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation. Even number exercises: 6.16 ) Find the area under the standard normal curve a) between z = 0 and z = 1.95 : => The area is .9744 - .5 = .4744 b) between z = .84 and z = 1.95 => the area is .9744 - .7995 = .1749 c) between z = 0 and z = -2.58 => the area is 0.5 - .0049 = .4951 d) between z = -.57 and z = -2.49 => the area is .2843 – .0064 = .2779 e) between z = - 2.15 and z = 1.87 => the area is .9693 - .0158 = .9535 6.18 ) Obtain the area under the standard normal curve : a) to the right of z = 1.43 : => the area is 1 - .9236 = .0764 b) to the right of z = -.65: => the area is 1 – 0.2578 = .7422 c) to the left of z = -1.65 : => the area is .0495 d) to the left of z = .89 : => the area is .8133 6.20 ) Obtain the area under the standard normal curve : a) from z = 0 to z = 3.94 : => the area is close to .5 b) between z = 0 and z= -5.16 : => the area is close to .5 c) to the right of z = 5.42 => the area is close to 1 d) to the left of z = -3.68: => => the area is close to 1 6.22) Determine the following probabilities for the standard normal distribution : a) P (-2.46 <= z <= 1.88 ) = .9699 - .0069 = .963 b) P ( 0 <= z <= 1.96) = .9750 - .5 = .475 c) P (-2.58 <= z <= 0) =.5 - .0049 = .4951 d) P ( z >= .73) = 1 - .7673 = 0.2327 6.24 ) Determine the following probabilities for the standard normal distribution : a) P ( z < -1.31 ) = .0951 approximately b) P ( 1.23 <= z <= 2.89 ) = .9981 – 0.8907 = .1074 c) P ( -2.24 <= z <= -1.19) = .1170 - .0125 = .1045 d) P (z < 2.02 ) = . 9783 approximately 6.26 ) Determine the following probabilities for the standard normal distribution : a) P ( z > -1.86 ) = 1- .0314 = .686 b) P ( -.68 <= z <= 1.94 ) = .9738 - .2483 = .7255 c) P ( 0 <= z <= 3.85 ) = 0.5 – 0.5 = 0 approximately d) P (-4.34 <= z <= 0) = 0.5 – 0.5 = 0 approximately. e) P ( z >4.82 ) is close to 0 f) P ( z < -6.12 ) is close to 0 MGT 218: Chapter 6: Continuous Random Variables and the Normal distribution (PART 2) with Solution for Exercises on Text book. Theory : Standardizing a Normal Distribution : Converting an x Value to a z Value : For a normal random variable x, a particular value of x can be converted to its corresponding z value by using the formula : where Mean and standard deviation are the mean and standard deviation of the normal distribution of x, respectively. When x follows a normal distribution , z follows the standard normal distribution. Practice Questions and Solutions on Textbook ( Introductory Statistics 8e edition : Prem S.Mann) – I think this will help because the answers are not provided on the textbook. 6.38 ) According to the U.S.Employment and Training Administration, the average weekly unemployment benefit paid out in 2008 was $297. Suppose that the current distribution of weekly unemployment benefits paid out is approximately normally distributed with a mean of $297 and a standard deviation of $74.42. Find the probability that a randomly selected American who is receiving unemployment benefits is receiving : a) more than $ 400 per week b) between $200 and $340 per week. We have : mean = $ 297 , standard deviation = 74.42 a) Z = 400 – 297 / 74.42 = 1.38 P ( x < 400 ) = .9162 b) z1 = 200 – 297 / 74.42 = -1.30 z2 = 340 – 297 / 74.42 = 0.58 p( 200 < x < 340 ) = .7190 - .0968 = .6222 6.40 ) Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. His times between pitches are normally distributed with a mean of 36 seconds and a standard deviation of 2.5 seconds. What percentage of his times between pitches are : (mean = 36 , standard deviation = 2.5) a) longer than 39 seconds : z= 39 – 36 / 2.5 = 1.2 P ( x < 39 ) = .8849 c) between 29 and 34 seconds : z1 = 29-36 / 2.5 = -2.8 z2 = 43 – 36 / 2.5 = 2.8 P ( 29 < x < 34 ) = .9974 - .0026 = .9948 6.42 ) The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balances on all Visa credit cards issued by the Bank of Connecticut have a mean of $ 850 and a standard deviation of $270 . Assume that the balances on all these Visa cards follow a normal distribution. a. What is the probability that a randomly selected Visa card issued by this bank has a balance between $1000 and $1440 ? b. What percentage of the Visa cards issued by this bank have a balance of $730 or more? a) z1 = 1000 – 850 / 270 = .56 z2 = 1440 – 850 / 270 = 2.19 P ( 1000 < x < 1440 ) = .9857 - .7123 = . 2734 b) z = 730 – 850 / 270 = -.44 P ( x >730 ) = 1 - .3300 = .6700 6.44) The transmissions on a model of a specific car has a warranty of 40,000 miles. It is known that the life of such transmissions has a normal distribution with a mean of 72,000 miles and a standard deviation of 12,000 miles. A transmission is chosen at random. a)What is the probability the transmission will fail before the end of the warranty period? b)What is the probability the transmission will be good for more than 100,000 miles? Mean = 72000 , standard deviation = 12000 a) z = 40000 – 72000 / 12000 = -2.67 P ( x< - 2.67 ) = .0038 b) z = 100000 – 72000 / 12000 = 2.33 P ( x > 100000 ) = .0158 6.46 ) The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at the supermarket. The expectation of the management is that after the promotional policy is advertised, the expenditures for all customers at the supermarket will be normally distributed with a mean of $95 and a standard deviation of $20. If the management wants to give free gift to all the customers who spend more than $130, what percentage of the customers are expected to get free gifts ? Z = 130 – 95 / 20 = 1.75 P ( x > 130 ) = 1 .9599 = .0401 6.48 ) A psychologist has devised a stress test for dental patients sitting in the waiting rooms. According to this test, the stress scores ( on a scale of 1 to 10 ) for patients waiting for root canal treatments are found to be approximately normally distributed with a mean of 7.59 and a standard deviation of .73 . a) What percentage of such patients have a stress score lower than 6.0 ? b) What is the probability that a randomly selected root canal patient sitting in the waiting room has a stress score between 7.0 and 8.0 ? c) The psychologist suggests that any patient with a stress score of 9.0 or higher should be given a sedative prior to treatment . what percentage of patients waiting for root canal treatments would need a sedative if this suggestion is accepted ? a) z = 6 – 7.59 / .73 = -2.18 => P ( x < 6 ) = .0146 b) z1 = 7 – 7.59 / .73 = -.81 z2 = 8 – 7.59 / .73 = .56 = > P (7 < x < 8 ) = .7123 - . 2090 = .5033 c) z = 9 – 7.59 / .73 = 1.93 P ( x > 9 ) = 1 - .9732 = .0268 6.50 ) Fast Auto Service guarantees that the maximum waiting time for its customers is 20 minutes for oil and lube service on their cars. It also guarantees that any customer who has to wait longer than 20 minutes for this service will receive a 50% discount on the charges. It is estimated that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. Suppose the time taken for oil and lube service on a car follows a normal distribution. a. What percentage of customers will receive a 50% discount on their charges? b. Is it possible that it may take longer than 25 minutes for oil and lube services? a) Z = 20 – 15 / 2.4 = 2.083 P ( x > 20 ) = 1 - .9977 = 0.0023 b) Z = 25 – 15 / 2.4 = 4.1677 > 4 It is impossible 6.58 ) Let x be a continuous random variable that follows a normal distribution with a mean of 550 and a standard deviation of 75. a) Find the value of x so that the area under the normal curve to the left of x is .0250 P(x) = .0250 => z= -1.96 X = -1.96 . 75 + 550 = 403 c) Find the value of x so that the area under the normal curve to the right of x is .9345 P (x) = 1- .9345 = .0655 => z = -1.51 X = -1.51 . 75 + 550 = 436.75 d) Find the value of x so that the area under the normal curve to the right of x is approximately 0.0275 P(x) = 1- .0275 = .9725 => z = 1.92 X = 1.92 . 75 + 550 = 694 e) Find the value of x so that the area under the normal curve to the left of x is approximately .9600 P(x) = .9600 => z= 1.75 X = 1.75 . 75 + 550 = 681.25 f) Find the value of x so that the are under the normal curve between mean and x is approximately .4700 and the value of x is less than mean. X is less than mean => P(x) < 0.5 P(x)= 0.5 – 0.4700 = .0030 => z = -2.75 X = -2.75 . 75 + 550 = 343.75 g) Find the value of x so that the area under the normal curve between mean and x is approximately .4100 and the value of x is greater than mean. X is greater than mean => P(x) > 0.5 P(x) = .5 + .41 = .91 => z = 1.34 X = 1.34 . 75 + 550 = 650.5 6.60 ) The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised , the expenditures for all customers at this supermarket will be normally distributed with a mean of $95 and a standard deviation of $20. If the management wants to give free gifts to at most 10% of the customers, what should be the amount be above which a customer would receive a free gift ? P(x) = 10% => z = -1.28 X = -1.28 . 20 + 95 = 69.4 Chapter 7 : Sampling Distributions Learning objectives : 1) Sampling distribution , sampling errors and non-sampling errors : 2) Mean and standard deviation of x bar 3) Shape of the sampling distribution of x bar 4) Applications of the sampling distribution of x bar 5) Population and sample proportions; and Mean, standard deviation and shape of the sampling distribution of p 6) Applications of the sampling distribution of p 7.1) Sampling distribution , sampling errors and nonsampling errors : * The population distribution is the probability distribution derived from the information on all elements of a population data. The central limit theorem states that if a large enough sample is taken (typically n > 30) then the sampling distribution of x¯x¯ is approximately a normal distribution with a mean of μ and a standard deviation of σn√σn. Since in practice we usually do not know μ or σ we estimate these by x¯x¯ and sn√sn respectively. In this case s is the estimate of σ and is the standard deviation of the sample. The expression sn√sn is known as the standard error of the mean, labeled SE(x¯x¯) Simulation: Generate 500 samples of size heights of 4 men. Assume the distribution of male heights is normal with mean μ = 70" and standard deviation σ = 3.0". Then find the mean of each of 500 samples of size 4. Here are the first 10 sample means: 70.4 72.0 72.3 69.9 70.5 70.0 70.5 68.1 69.2 71.8 Theory says that the mean of ( x¯x¯ ) = μ = 70 which is also the Population Mean and SE(x¯)=σn√=34√=1.50SE(x¯)=σn=34=1.50 Simulation shows: Average (500 x¯x¯'s) = 69.957 and SE(of 500 x¯x¯'s) = 1.496 Change the sample size from n = 4 to n = 25 and get descriptive statistics: Theory says that the mean of ( x¯x¯) = μ = 70 which is also the Population Mean and SE(x¯)=σn√=325√=0.60SE(x¯)=σn=325=0.60 Simulation shows: Average (500 x¯x¯'s) = 69.983 and SE(of 500 x¯x¯'s) = 0.592 Sampling Distribution of Sample Mean x¯x¯ from a Non-Normal Population Simulation: Below is a Histogram of Number of Cds Owned by PSU Students. The distribution is strongly skewed to the right. Assume the Population Mean Number of CDs owned is μ = 84 and σ = 96 Let's obtain 500 samples of size 4 from this population and look at the distribution of the 500 x-bars: Theory says that the mean of ( x¯x¯) = μ = 84 which is also the Population Mean the SE(x¯)=48=964√SE(x¯)=48=964 Simulation shows Average(500 x¯x¯'s) = 81.11 and SE(500 x¯x¯'s for samples of size 4) = 45.1 Change the sample size from n = 4 to n = 25 and get descriptive statistics and curve: Theory says that the mean of ( x¯x¯) = μ = 84 which is also the Population Mean and the SE(x¯)=9625√=19.2SE(x¯)=9625=19.2 Simulation shows Average(500 x¯x¯'s) = 83.281 and SE(500 x¯x¯'s for samples of size 25) = 18.268. A histogram of the 500 x¯x¯'s computed from samples of size 25 is beginning to look a lot like a normal curve. i. The Law of Large Numbers says that as the sample size increases the sample mean will approach the population mean. ii. The Central Limit Theorem says that as the sample size increases the sampling distribution of X¯X¯ (read x-bar) approaches the normal distribution. We see this effect here for n = 25. Generally, we assume that a sample size of n = 30 is sufficient to get an approximate normal distribution for the distribution of the sample mean. iii. The Central Limit Theorem is important because it enables us to calculate probabilities about sample means. Example. Find the approximate probability that the average number of CDs owned when 100 students are asked is between 70 and 90. Solution. Since the sample size is greater than 30, we assume the sampling distribution of x¯x¯ is about normal with mean μ = 84 and SE(x¯)=σn√=96100√=9.6SE(x¯)=σn=96100=9.6. We are asked to find Prob( 70 < X¯X¯ < 90). The z-scores for the two values are for 90: z = (90 - 84)/ 9.6 = 0.625 and for 70: z = (70-84)/9.6 = -1.46. From tables of the normal distribution we get P( -1.46 < Z < 0.625) = .734 - .072 = .662. Suppose the sample size was 1600 instead of 100. Then the distribution of x¯x¯ would be about normal with mean 84 and standard deviation σn√=961600√=9640=2.4σn=961600=9640=2.4. From the empirical rule we know that almost all x-bars for samples of size 1600 will be in the interval 84 ± (3)(2.4) or in the interval 84 ± 7.2 or between 76.8 and 91.2. The Law of Large Numbers says that as we increase the sample size the probability that the sample mean approaches the population mean is 1.00! MGT 218 : Statistics questions for chapter 7 : Sample Distributions Chapter 8 : Estimation , Point Estimate and Internal Estimate 1) Theory : a) Estimation : The assignment of value(s) to a population parameter based on a value of the corresponding sample statistics is called estimation. b) Estimate and Estimator : The value(s) assigned to a population parameter based on the value of a sample statistics is called an estimate. The sample statistic used to estimate a population parameter is called an estimator. c) Point estimate : the value of a sample statistic that is used to estimate a population parameter is called a point estimate. d) Interval estimation : in interval estimation, an interval is constructed around the point estimate, and it is stated that this interval is likely to contain the corresponding population parameter. e) Confidence level and confidence interval : Each interval is constructed with regard to a given confidence level and is called a confidence interval. The confidence interval is given as Point estimate ( + - ) Margin of error The confidence level associated with a confidence interval states how much confidence we have that this interval contains the true population parameter. The confidence level is denoted by ( 1 –a)100% f) Estimation of a population mean : standard deviation known Confidence Interval for μ : the (1-a)100% confidence interval for μ under Cases I and II is The value of z used here is obtained from the standard normal distribution table ( Table IV of Appendix C) for the given confidence level. Margin of Error : for the estimate for μ, denoted by E, is the quantity that is subtracted from and added to the value of x bar to obtain a confidence interval for μ , thus With Case 1 . If the following three conditions are fulfilled : 1- The population standard deviation σ is known. 2- The sample size is small ( n < 30 ) 3- The population from which the sample is selected is normally distributed. Case 2. If the following two conditions are fulfilled : 1- The population standard deviation σ is known 2- The sample size is large ( n >=30 ) n/N <= 0.05 Case 3. If the following three conditions are fulfilled 1. The population standard deviation σ is known. 2. The sample size is small ( n < 30 ) 3. The population from which the sample is selected is not normally distributed ( or its distribution is unknown ). 2) Exercises from the textbook : 7.4 ) a ) find μ = 15 + 13 +8+17+9+12/6 = 12.33 b) xx̄ = 13+8+9+12 / 4 = 10.2 Sampling error = 10.2 – 12.33 = -2.13 c) xx̄ = 13+8+6+12 / 4= 9.75 Nonsampling error = 9.75 – 10.2 = -0.45 d) Numbers : 17,9,12,8 xx̄ = (17+9+12+8)/4 = 11.5 Sampling error = 11.5 – 12.33 = -.83 7.8 ) The following data give the years of teaching experience for all ive faculty members of a department at a university. 7 8 14 7 20 a) Let x denote the years of teaching experience for a faculty member of this department. Write the population distribution of x b) List all the possible samples of size three ( without replacement ) that can be selected from this population. Calculate the mean for each of these samples. Write the sampling distribution of xx̄ c) Calculate the mean for the population data. Select one random sample of size three and calculate the sample mean xx̄. Compute the sampling error. a) Population probability distribution : X P(x) 7 .40 8 .20 14 .20 20 .20 = 1.00 b) Sample xx̄ F P(x) Sampling error 7,8,14 9.67 2 .2 -1.53 7,8,7 7.33 1 .1 -3.87 7,8,20 11.67 2 .2 0.47 8,14,20 14 1 .1 2.8 14,7,20 13.67 2 .2 2.27 7,14,7 9.33 1 .1 -1.87 7,7,20 11.33 1 .1 0.13 = 1 With μ = 11.2 7.14 ) Consider a large population with μ= 90 and σ=18. Assuming n/N <= .05, find the mean and standard deviation of sample mean for a sample size of : a) n= 18 => σ =M2.36 μ (x)= 60 b) n = 90 => σ M = 21.21 μ (x)= 60 7.16) A population of N = 100000 has σ = 40. In each of the following cases, which formula will you use to calculate σx for each of these cases. a) n = 2500 => n/N = 0.025 < 0.05 => σx = 0.8 b) n = 7000 => n/N = 0.357 < 0.05 => σx = 0.46 7.18 ) For a population , μ = 46 and σ = 10. a) For a sample selected from this population, μx = 46 and σx =2.0. Find the sample size. Assume n/N <= .05 => n = 25 b) n = 39.0625 7.20 ) The living spaces of all homes in a city have a mean of 2300 square feet and a standard deviation of 500 square feet. Let xx̄ be the mean living space for a random sample of 25 homes selected from this city. Find the mean and standard deviation of the sampling distribution of xx̄. Mean of sample = mean of population σx = 100 7.24) The standard deviation of the 2011 gross sales of all corporations is known to be $139.50 million. Let xx̄ be the mean of the 2011 gross sales of a sample of corporations. What sample size will produce the σx to $15.50 million? Assume n/N <=0.05 => n^2 = 139.50^2 / 15.50^2 = 81 8.24 ) A city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 40 households from the city, which gave the mean water usage to be 3415.70 gallons over a one-month period. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is 389.60 gallons. Make a 95 % confidence interval for the average monthly residential water usage for all households in this city. 95% confidence interval : => z = 1.96 3415.70 ( +- ) 1.96 x ( 389.60 / 6.32 ) : the range is from 3294.96 to 3536.44 8.26) Lazurus Steel Corporation products iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods vary slightly. It is known that when the machine is working properly, the mean length of the rods made on this machine is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to .10 inch. The quality control department takes a sample of 20 such rods every week, calculates the mean length of these rods, and makes a 99% confidence interval for the population mean. If either the upper limit of this confidence interval is greater than 36.05 inches or the lower limit of this confidence interval is less than 35.96 inches, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of 36.02 inches. Based on this sample, will you conclude that the machine needs an adjustments? Assume that the lengths of all such rods have a normal distribution. Solution : We have xx̄ = 36.02 and n = 20 with 99% confidence interval => z = 2.58 The range is from 36.02 (+-) 2.58 x (0.1 / 4.472) = 35.96 to 36.08. The machine needs adjustment. 8.44) Find the value of t from the t distribution table for each of the following : a) Confidence level = 99% and df = 13 => t = 3.012 b) Confidence level = 95% and n=36 => df = 35 => t= 2.030 c) Confidence level = 90% and df = 16 => t= 1.746 8.28 ) A consumer agency that proposes that lawyer’s rates are too high wanted to estimate the man hourly rate for all lawyers in New York city. A sample of 70 lawyers taken from New York city showed that the mean hourly rate charged by them is $570. The population standard deviation of hourly charges for all lawyers in New York city is $110. a) Construct a 99% confidence interval for the mean hourly charges for all lawyers in NY city. b) Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best ? n = 70 , xx̄ = 570 , σ = 110 : a) With 99% => z = 2.58 : The range is from 570 (+-) 2.58 x 110 / 8.367 = 536.08 to 603.92. 8.30 ) A marketing researcher wants to find a 95% confidence interval for the mean amount that visitors to a theme park spend per person per day. She knows that the standard deviation of the amounts spent per person per day by all visitors to this park is $11 . How large a sample should the researcher select so that the estimate will be within $2 of the population mean? … E = 2 , σ = 11 , 95 % => z = 1.96 n = 116.2084 . => round up to 117. 8.32) A department store manager wants to estimate at a 98% confidence level the mean amount spent by all customers at this store. From an earlier study, the manager knows that the standard deviation of amounts spent by customers at this store is $31. What sample size should he choose so that the estimate is within 3$ of the population mean? 98% => z = 2.33 n = 579.69 => round up to 580 8.34) You are interested in estimating the mean commuting time from home to school for all commuter students at your school. Briefly explain the procedure you will follow to conduct this study. Collect the required data from a sample of 30 or more such students and then estimate the population mean at a 99% confidence level. Assume that the population standard deviation for such times is 5.5 minutes. 99 % => z = 2.58 , n= 30 , σ = 5.5 . => E^2 = (5.5^2 x 2.58^2) / 30 = 6.712 => E = 2.59.
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