simple assignment

Anonymous
timer Asked: Feb 7th, 2019
account_balance_wallet $15

Question Description

Note1: Print off a copy of this Microsoft Word file and answer each question directly on your hard paper copy of these pages.

Note2: Hand in your answer to Q#1, Q#2, and Q#3 only.

Note 3: This assignment does NOT cover all the material that you are responsible for knowing in this course. You are, of course, responsible for the material covered in all readings, lectures, assignments, and practice problems


please use my Formula table.


Unformatted Attachment Preview

MGSC2207 Winter 2019 MGSC2207 ASSIGNMENT #4 due 11:00 am 8 Feb 2019 nd (Assignment drop boxes Loyola 2 floor) Section _________________________________________ Name (print) ____________________________________ Signature _______________________________________ Student ID# _____________________________________ Note1: Print off a copy of this Microsoft Word file and answer each question directly on your hard paper copy of these pages. Note2: Hand in your answer to Q#1, Q#2, and Q#3 only. Note 3: This assignment does NOT cover all the material that you are responsible for knowing in this course. You are, of course, responsible for the material covered in all readings, lectures, assignments, and practice problems (on the P: drive). Total marks for this assignment = 38 1. A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $33.77 and a standard deviation of $15. a) Develop a 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share. 6 marks b) Explain what the interval you found tells you. 2 marks Page 1 of 6 MGSC2207 Winter 2019 c) What sample size would be necessary to achieve a margin of error of  $2? Use a confidence level of 95%. 6 marks d) Three years ago the mean price charged for a trade of 100 shares at $50 per share was $39.25. Has the price dropped significantly? Justify. 2 marks Page 2 of 6 MGSC2207 Winter 2019 2. Workers in the health care industry were surveyed to determine the proportion of workers who feel their industry is understaffed. 33% of the workers surveyed said they were understaffed (USA Today, January 11, 2010). We do not know how many workers had been surveyed, but we will assume it to be 200. a) Construct a 95% confidence interval for the proportion of workers in the health care industry who feel their industry is understaffed. Interpret the meaning of your confidence interval in plain non-technical language. 8 marks b) How large would the sample need to be to ensure that the margin of error is  5% or less for each of the three confidence intervals? Perform the calculation using an appropriate pilot study proportion as well as a worst case scenario. 6 marks Page 3 of 6 MGSC2207 Winter 2019 3. HOMEGROCER.com is an on-line grocery store in the Seattle area that has more than 10,000 customers. The following is a shopping list of eight items from HomeGrocer.com and local Seattle supermarkets. Products Tide High Efficiency, 64 oz. Oreo Cookies, 20 oz. Formula 409 Cleaner, 22 oz. Pampers Newborn Diapers, 40 count Coke Classic, dozen 12 oz. Cans Colgate Total Toothpaste, 7.8 oz. Tropicana Orange Juice, 64 oz. Cheerrios Whole Grain Cereal, 20 oz. HomeGrocer Supermarkets 6.99 6.99 3.29 3.49 2.59 2.69 10.79 3.99 3.49 3.59 10.99 3.59 3.49 3.49 4.29 3.99 a) Construct a 95% confidence interval estimate of the difference in the average price for products purchased from HOMEGROCER.com and Seattle supermarkets. 6 marks b) Using your confidence interval results, is there evidence of a significant difference in average price for the basket of goods? 2 marks Page 4 of 6 MGSC2207 Winter 2019 Do NOT hand in your answer to the following questions!! 4. The R&M department store has two charge plans available for its credit-account customers. The management of the store wishes to collect information about each plan and to study the differences between the two plans. It is interested in the average monthly balance. Random samples of 25 accounts of plan A and 50 accounts of plan B are selected with the following results: Plan A Plan B Size 25 50 Mean $75 $110 Standard deviation $15 $14.14 Is there evidence of a difference in the average monthly balances between plan A and plan B? Construct a 95% confidence interval to answer this question. Include a statement. Page 5 of 6 MGSC2207 Winter 2019 5. A survey conducted by the Canadian Automobile Association (CAA) showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $74.50. a) Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls. Include a statement explaining your interval. b) What role, if any, did the Central Limit Theorem play in your calculations? Justify. c) Based on the confidence interval, does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the CAA? Explain. Page 6 of 6 Standard Normal Distribution 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 t distribution One-tail α two-tail α df 0.4 0.8 0.25 0.5 0.1 0.2 1.00 0.325 1.000 3.078 6.314 12.706 2.00 0.289 0.816 1.886 2.920 3.00 0.277 0.765 1.638 2.353 4.00 0.271 0.741 1.533 5.00 0.267 0.727 6.00 0.265 7.00 0.005 0.01 0.001 0.002 0.0005 0.001 31.821 63.657 318.310 636.619 4.303 6.965 9.925 22.326 31.599 3.182 4.541 5.841 10.213 12.924 2.132 2.776 3.747 4.604 7.173 8.610 1.476 2.015 2.571 3.365 4.032 5.893 6.869 0.718 1.440 1.943 2.447 3.143 3.707 5.208 5.959 0.263 0.711 1.415 1.895 2.365 2.998 3.499 4.785 5.408 8.00 0.262 0.706 1.397 1.860 2.306 2.896 3.355 4.501 5.041 9.00 0.261 0.703 1.383 1.833 2.262 2.821 3.250 4.297 4.781 10.00 0.260 0.700 1.372 1.812 2.228 2.764 3.169 4.144 4.587 11.00 0.260 0.697 1.363 1.796 2.201 2.718 3.106 4.025 4.437 12.00 0.259 0.695 1.356 1.782 2.179 2.681 3.055 3.930 4.318 13.00 0.259 0.694 1.350 1.771 2.160 2.650 3.012 3.852 4.221 14.00 0.258 0.692 1.345 1.761 2.145 2.624 2.977 3.787 4.141 15.00 0.258 0.691 1.341 1.753 2.131 2.602 2.947 3.733 4.073 16.00 0.258 0.690 1.337 1.746 2.120 2.583 2.921 3.686 4.015 17.00 0.257 0.689 1.333 1.740 2.110 2.567 2.898 3.646 3.965 18.00 0.257 0.688 1.330 1.734 2.101 2.552 2.878 3.610 3.922 19.00 0.257 0.688 1.328 1.729 2.093 2.539 2.861 3.579 3.883 20.00 0.257 0.687 1.325 1.725 2.086 2.528 2.845 3.552 3.850 21.00 0.257 0.686 1.323 1.721 2.080 2.518 2.831 3.527 3.819 22.00 0.256 0.686 1.321 1.717 2.074 2.508 2.819 3.505 3.792 23.00 0.256 0.685 1.319 1.714 2.069 2.500 2.807 3.485 3.768 24.00 0.256 0.685 1.318 1.711 2.064 2.492 2.797 3.467 3.745 25.00 0.256 0.684 1.316 1.708 2.060 2.485 2.787 3.450 3.725 26.00 0.256 0.684 1.315 1.706 2.056 2.479 2.779 3.435 3.707 27.00 0.256 0.684 1.314 1.703 2.052 2.473 2.771 3.421 3.690 28.00 0.256 0.683 1.313 1.701 2.048 2.467 2.763 3.408 3.674 29.00 0.256 0.683 1.311 1.699 2.045 2.462 2.756 3.396 3.659 30.00 0.256 0.683 1.310 1.697 2.042 2.457 2.750 3.385 3.646 40.00 0.681 1.303 1.684 2.021 2.423 2.704 3.307 3.551 50.00 0.679 1.299 1.676 2.009 2.403 2.678 3.261 3.496 60.00 0.679 1.296 1.671 2.000 2.390 2.660 3.232 3.460 90.00 0.677 1.291 1.662 1.987 2.369 2.632 3.185 3.403 1.289 1.658 1.980 2.358 2.617 3.160 3.373 1.282 1.645 1.960 2.326 2.576 3.090 3.291 120.00 inf 0.253 0.674 0.05 0.025 0.01 0.1 0.05 0.02 Critical values ofstudent's t MGSC2207 LEE EQUATIONS All equations on this page (except #6) are slight variations of just one equation (a) which have been designed to handle different real-life situations. In each row below are two variations of exactly the same equation (i.e., statistical test). a σ known Parametric statistical tests σ unknown one-sample t-test for means  s  1  X  t    n indep-groups t-test for diff b/t means 2 1 −  2 (X1 − X 2 )  t s2 s2 + n1 n2 b σ known z = Jan 2019 X−  1 - 2  s d  t d  n  d     n Tests of Means 1 t 2 s (X t = SS X = X X− = n 1 - X 2 ) − (1 −  2 ) s2 s2 + n1 n2 dep-groups t-test for diff b/t means 3 c σ unknown z = 3 t = d − d sd p1 q1 p 2 q 2 + n1 n2 s= (n1 − 1) s12 + (n 2 − 1) s 22 n1 + n 2 − 2 df = n - 1 = ( weighted ) avg std dev df = n1 + n2 - 2 SS d sd = nd − 1 (d ) 2 SS d = d − nd 2 nd 5a if H 0 : has the form  1 =  2 ( p1 − p 2 ) − ( 1 −  2 ) pq pq + n1 n2 d d = X1 − X 2 total no. of successes total sample size = weighted avg p p n + p 2 n2 = 1 1 n1 + n2 p= 5b if H 0 : has the form  1 =  2 + ? % ( p1 − p 2 ) − ( 1 −  2 ) z = Non-parametric statistical tests (for testing weak numbers) (no σ – there is no such thing as a population std dev for weak numbers) n Tests of Proportions (Binomial) p − 4 z=  (1 −  ) n z = 5  1 −  2 ( p1 − p 2 )  z n (X )2 SS X n −1 nd df = nd – 1 two-group z-test for diff b/t props − s s = d= one-sample z-test for prop pq pz 4  n 2 X− 6  p1 q1 p 2 q 2 + n1 n2 2 (O − E ) 2 =  E dfchisquare = (r – 1) (c – 1) multiply m arg inals of cell E cell = grand total Correlation/Regression r= SS XY SS X SSY n = number of cases k = number of predictors ρ = population correlation (this is the Greek letter rho, not a “p”)  Y = b0 + b1 X Y b1 −  1 sb1 t SLOPE = for any k SS ERROR = SSY − b1 * SS XY Y = s ERROR = SS ERROR n − k −1 1 SS X k = number of predictors e 1 ( X − X )2 + n SS X  t se 1 + ANOVA FEffect = e s for any k  Y ts s for any k df = n – 2 SS XY SS X  YX 1− r2 n−2 df = n – 2 b0 = Y − b1 X b1 = r− t COR = FREG = 1 (X − X )2 + n SS X MS EFFECT MS ERROR To compute the total variation or SSTOTAL, pretend that all N scores are in a single group and compute the variation there is among all N scores. In independent groups or BG designs, the error variation is the variation within groups SSWG. Any variation within a column is always error variation. Why? Because all individuals in a column received the same treatment and should, therefore, be identical. Adding up the variation in each of the treatment groups (i.e., within each column) gives the SSWG. In WG or RM designs, the error variation is the residual error variation SSRESIDUAL. Multiple comparison test q = studentized range statistic value SS ( X ) For BG variation, make all scores in the first column equal to mean for that column. Do the same for each column. These scores have no variation within a group so the total variation among these scores must equal the variation between groups SSBG. Compute the SSBG using the equation above for SSTOTAL on these new scores (this new total variation contains no error variation and is, therefore, between-group variation only). An alternative is to use the equation below to compute the SSBG. = n( X1 − X ) 2 + n( X 2 − X ) 2 + ... df BG = k −1 grand mean (i.e., mean of all scores) Min diff = q * TOTAL =  (Y − Y ) 2 Differences between Means: Dep Groups (or RM) df TOTAL = N −1 = SS1 + SS 2 +... SSk df W G = k (n − 1) For BG variation, see comment to left. Alternatively, SS BG = n( X1 − X ) 2 + n( X 2 − X ) 2 + ... df BG = k −1 Compute variation due to individual differences by making all scores in a row = mean for individual in that row (or use eqn below). Compute SSIND DIFFS using either the equation for SSTOTAL on these new scores or use equation below. SS IND DIFFS = k ( X A − X ) 2 + k ( X B − X ) 2 + ... df IND DIFFS = n −1 SS RESIDUAL = SSWG − SS IND DIFFS df RESIDUAL = ( k − 1)(n − 1) SS TOTAL = SS BG + SS IND DIFFS + SS RESIDUAL where N = total no. of scores n = no. of individuals k = no. of scores per person where N = total no. of scores n = no. of individuals k = no. of groups X= =  (Y − Y ) 2 RESIDUAL 2 = SS1 + SS 2 +... + SSk df WG = k (n − 1) SS TOTAL = SS BG + SSWG BG  SS SS N df TOTAL = N −1 SS  df REGRESSION = k df RESIDUAL = n − k − 1 =X2 − TOTAL = se 2 SS REGRESSION =  (Y − Y ) MS REGRESSION MS RESIDUALl Differences between Means: Indep Groups b1 MS W G ERROR n X= grand mean (i.e., mean of all scores) Min diff = q * MS RESIDUAL ERROR n ...
Purchase answer to see full attachment

Tutor Answer

mickeygabz
School: UC Berkeley

See attached

MGSC2207 Winter 2019

MGSC2207

ASSIGNMENT #4

due 11:00 am 8 Feb 2019

nd

(Assignment drop boxes Loyola 2 floor)
Section _________________________________________
Name (print) ____________________________________
Signature _______________________________________
Student ID# _____________________________________
Note1: Print off a copy of this Microsoft Word file and answer each question directly on your hard
paper copy of these pages.
Note2: Hand in your answer to Q#1, Q#2, and Q#3 only.
Note 3: This assignment does NOT cover all the material that you are responsible for knowing in this
course. You are, of course, responsible for the material covered in all readings, lectures,
assignments, and practice problems (on the P: drive).

Total marks for this assignment = 38

1. A sample survey of 54 discount brokers showed that the mean price charged for a trade of
100 shares at $50 per share was $33.77 and a standard deviation of $15.
a) Develop a 95% confidence interval for the mean price charged by discount brokers for
a trade of 100 shares at $50 per share.
6 marks
Confidence interval = sample mean ± z critical * σ/√𝑛
Z critical at 95% confidence level = 1.96
Confidence interval = 33.77 ± 1.96 * 15/√54
=33.77 ± 4
= (29.77, 37.77)

b) Explain what the interval you found tells...

flag Report DMCA
Review

Anonymous
awesome work thanks

Similar Questions
Hot Questions
Related Tags
Study Guides

Brown University





1271 Tutors

California Institute of Technology




2131 Tutors

Carnegie Mellon University




982 Tutors

Columbia University





1256 Tutors

Dartmouth University





2113 Tutors

Emory University





2279 Tutors

Harvard University





599 Tutors

Massachusetts Institute of Technology



2319 Tutors

New York University





1645 Tutors

Notre Dam University





1911 Tutors

Oklahoma University





2122 Tutors

Pennsylvania State University





932 Tutors

Princeton University





1211 Tutors

Stanford University





983 Tutors

University of California





1282 Tutors

Oxford University





123 Tutors

Yale University





2325 Tutors