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YUCM_STAT311_161_1_FinalMA_to_Publish

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ROYAL COMMISSION YANBU COLLEGES AND INSTITUTES
YANBU UNIVERSITY COLLEGE
DEPARTMENT OF GENERAL STUDIES
STAT311-M (Statistics for Management II)
FINAL EXAMINATION SEMESTER I
1437–1438(H)/2016–2017(G)

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YUCM_STAT311_161_1_FinalMA_to_Publish

  1. 1. ROYAL COMMISSION YANBU COLLEGES AND INSTITUTES YANBU UNIVERSITY COLLEGE DEPARTMENT OF GENERAL STUDIES ! FINAL EXAMINATION SEMESTER I, 1437 – 1438 (H) / 2016 – 2017 (G) PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY 1. Answer all questions. 2. Do not use pencil for answering. 3. Write your ID number on the question paper. 4. Keep your mobiles switched off during the examination. 5. Read each question carefully before answering the questions. 6. You may use only non-programmable calculator. 7. Borrowing of calculator is not allowed. 8. No support material written or printed is allowed in the examination hall/room. 9. Number shown on the question paper against each question is the mark allocated to that question. Evaluated by: ___________________ Signature: ________________ Date: ____________ Reviewed by: ___________________ Signature: ________________ Date: ____________ Student’s Name: _____________________________________ Student’s ID: _____________ Course: _____STAT311-M______ Section: _____________ Date: 10-Jan-2017 (Tuesday) Q. No. Marks Assigned Marks Obtained Course Learning Outcomes CLOs Initials 1 5 1.2, 2.2, 2.3, 2.12 2 5 1.1, 2.2, 2.4, 2.12 3 10 1.2, 1.3, 2.2, 2.5, 2.12 4 10 1.2, 2.1, 1.7, 2.7,2.12 5 10 1.4, 1.5, 2.8, 2.9, 2.12 6 5 1.3,1.6, 1.8, 1.9, 2.1 7 5 1.9, 2.11 Total 50 MODEL ANSWER
  2. 2. A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers in a large school district. Using a significance level of α = 0.05, test the hypothesis that: . To test the hypothesis, Assume random samples have been selected from the two normally distributed populations with equal variances. The following sample data were observed: Solution Q1 : - Since the population standard deviations are unknown and must be estimated from the sample data, the test statistic will be a t-value from the t distribution - The critical value comes from a t-distribution with - The critical value from the t-distribution table is = 2.009. - When the population variances are assumed to be equal, using : - Since t = 1.2515 < 2.009, so the decision is to do not reject H0 at 0.05. - Based on the sample data, we conclude that There is not enough evidence to support the claim that the mean salary for the elementary school teachers is greater than the mean salary of the secondary school teachers. Q1 : Marks : 5 H0 :µ1 ≤ µ2 , HA :µ1 > µ2( ) (Use: t 0.025,50( ) = 2.009) salaries of elementary school teachers salaries of secondary school teachers x̅1 = $48200 x̅2 = $46600 s1 = $3500 s2 = $5500 n1 = 26 n2 = 26 df = n1 + n2 − 2 = 26 + 26 − 2 = 50 sp = (n1 −1)s1 2 +(n2 −1)s2 2 n1 + n2 −2 = (26−1)(3500)2 +(26−1)(5500)2 26+26−2 ≈ 4609.77 t = (x1 − x2 )−(µ1 −µ2 ) sp 1 n1 + 1 n2 = (48200−46600)−0 (4609.77) 1 26 + 1 26 ≈ 1.2515 Page of2 7 ROYAL COMMISSION YANBU COLLEGES AND INSTITUTES YANBU UNIVERSITY COLLEGE DEPARTMENT OF GENERAL STUDIES STAT311-M (Statistics for Management II) FINAL EXAMINATION SEMESTER I, 1437 – 1438 (H) / 2016 – 2017 (G) Time Allowed: 3 Hour Student’s ID: _____________________________________ Max. Marks: 50 1 2 1 1 2 1 2
  3. 3. Find the 90% confidence interval for the variance of the ages of seniors at Yanbu University College if a sample of 24 students has a standard deviation of 2.3 years. Assume the variable is normally distributed. Solution Q2 : For a 90% confidence interval we find the following values for and with n = 22 The confidence interval is calculated using Equation: ! Q2 : Marks : 5 Use : χ(0.05,23) 2 = 35.172 , χ(0.95,23) 2 =13.091( ) Hence, df = 24-1= 23 , α = 0.10 ⇒ α / 2( )= 0.05 ⇒ χ(0.05,23) 2 = 35.172 ⇒ 1−(α / 2)( )= 0.95 ⇒ χ(0.95,23) 2 =13.091 s =16 ⇒ s2 = 2.32 = 5.29 (n−1)s2 χU 2 ≤σ 2 ≤ (n−1)s2 χL 2 (24−1)2.32 35.172 ≤σ 2 ≤ (24−1)2.32 13.091 3.459 ≤σ 2 ≤ 9.294 Page of3 7 1 2 1 2 1 1 1 1
  4. 4. The average local cell phone monthly bill is $50. A random sample of monthly bills from three different providers is listed below. At α = 0.05 is there a difference in mean bill amounts among providers? The appropriate null and alternative hypotheses are : H0: µ1 = µ2 = µ3 , HA: At least one mean differs from the others (claim). Conduct the appropriate hypothesis test. 
 Solution Q3 : ! , ! , ! ! ,! H0: µ1 = µ2 = µ3 , HA: At least one mean differs from the others (claim). C.V. = ; d.f.N. = 3 -1 = 2; d.f.D. = 9 - 3 = 6 ! ! ! - Since the test statistic FRatio = 0.10 < 5.14, do not reject the null hypothesis. - There is not enough evidence to support the claim that at least one mean differs from the others. Q3 : Marks : 10 STC Mobily Zain 80 105 90 60 55 85 110 65 55 (Use : F0.05, 2, 6( ) = 5.14) X1 2 X1 -X( ) 2 X2 2 X2 -X( ) 2 X3 2 X3 -X( ) 2 STC Mobily Zain 80 6400 3 105 11025 711 90 8100 136 60 3600 336 55 3025 544 85 7225 44 110 12100 1003 65 4225 178 55 3025 544 250 22100 1342 225 18275 1433 230 18350 725 X1 = 250 3 ≈ 83.33 X2 = 225 3 = 75 X3 = 230 3 ≈ 76.67 X = xij i=1 k ∑ ⎛ ⎝⎜ ⎞ ⎠⎟ nT = 83.33 + 75 + 76.67 3 ≈ 78.33 X = xij j=1 ni ∑ i=1 k ∑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nT = 250 + 225 + 230 9 ≈ 78.33 F0.05, 2, 6( ) = 5.14 SST = x2 ij j=1 ni ∑ i=1 k ∑ − xij j=1 ni ∑ i=1 k ∑ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ 2 nT = 22100+18275+18350( )− (250+225+230)2 9 = 3500 = xij − x( )j=1 ni ∑ 2 i=1 k ∑ = 1342( )+ 1433( )+ 725( ) SSB = xij j=1 ni ∑ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ i=1 k ∑ n 2 − xij j=1 ni ∑ i=1 k ∑ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ 2 nT = (250)2 +(225)2 +(230)2 3 − (250+225+230)2 9 ≈ 116.67 ≈ ni xi − x( ) 2 i=1 k ∑ SSW = SST − SSB = 3500 −116.67 = 3383.33 ANOVA Source of Variation SS df MS F - Ratio Between Groups 116.67 2 58.34 0.10 Within Groups 3383.33 6 563.89 Total 3500 8 Page of4 7 1 1 1 1 3 3
  5. 5. To test the effectiveness of a new drug, a researcher gives one group of individuals the new drug and another group a placebo. The results of the study are shown here. At α = 0.10, can the researcher conclude that the drug is effective? (a) State the relevant null and alternative hypotheses. (b) Conduct the appropriate test and state a conclusion. 
 Solution Q4 : (a) The null and alternative hypotheses are: H0 : Medication is independent of effective (The drug is not effective) HA :There is a relationship between Medication and effective (The drug is effective). [claim] (b) Contingency Table : The expected cell frequencies are:
 ∴ Chi-square (χ 2 ) = 10.642 - The critical value of the test statistic: The degrees of freedom is = (2 -1) (2 -1) = 1. From the chi-square table α = 0,10 and 1 degrees of freedom, the critical value is χ 2 = 2,706. Thus, if the test statistic > 2,706, reject the null hypothesis. Otherwise, do not reject. - Make the decision: , we reject the null hypothesis. - There is enough evidence to support the claim that the drug is effective. Q4 : Marks :10 Medication Effective Not effective Drug 32 9 Placebo 12 18 (Use : χ 0.10( ) 2 = 2.706) o11 = o12 = e11 = e12 = o21 = o22 = e21 = e22 = Medication Effective Not effective Drug 32 9 41 25.41 15.59 Placebo 12 18 30 18.59 11.41 44 27 71 oij∑ = eij∑ = x2 = 32 25.41 6.59 43.45 1.71 9 15.59 -6.59 43.45 2.79 12 18.59 -6.59 43.45 2.34 18 11.41 6.59 43.45 3.81 71 71 10.642 oij (oij −eij )eij =e22 =o22 =e11 =o11 (oij − eij )2 =e12 =o12 (oij −eij )2 eij =o21 =e21 χCal 2 = 10.642 > χ(0.10, 1) 2 = 2,706 Page of5 7 5 1 1 2 1 2 2 1 2 1 2
  6. 6. The number of faculty and the number of students are shown for a random selection of departments in Yanbu University College. (a) Compute the correlation coefficient for these sample data. (b) Is there a significant relationship between the two variables? Test the hypothesis : . 
 Use a significance level of α = 0.05 (c) Compute the regression equation based on these sample data and interpret the regression coefficients. Solution Q5 : (a) - From the scatter plot it appears that a strong positive linear relationship exists between the two variables. (c) The test statistic is computed as follows: Or - Because t = 4.066 > +3.182, [critical t, with α/2 = 0.025, df = 5 - 2 = 3], reject the H0. There is sufficient evidence to conclude that the population correlation coefficient does not equal 0. and There is a significant linear relationship between the number of faculty and the number of students at departments in YUC. (d) the regression coefficients: 
 Q5 : Marks :10 Faculty 10 14 16 12 20 Students 120 130 170 150 200 [H0 : ρ = 0, HA : ρ ≠ 0] (t 0.025, 3( ) = +3.182) x2 y2 xy (x − x)2 (y− y)2 (x − x)(y− y)x y 10 120 100 14400 1200 19 1156 150 14 130 196 16900 1820 0 576 9.60 16 170 256 28900 2720 3 256 25.60 12 150 144 22500 1800 6 16 10 20 200 400 40000 4000 31 2116 257.60 72 770 1096 122700 11540 59 4120 452 0.9152 r = n xy − x y∑∑∑ [n (x2 )− ( x)2 ][n (y2 )− ( y)2 ∑∑ ]∑∑ = 5(11540)− 72 × 770 5(1096)− (72)2 ⎡⎣ ⎤⎦ 5(122700)− (770)2 ⎡⎣ ⎤⎦ ≈ 0.92 r = (x − x)(y − y)∑ [ (x − x)2 ][ (y − y)2 ]∑∑ = 452 59[ ] 4120[ ] ≈ 0.915 ∵r = 0.92, n = 5, H0 : ρ = 0, HA : ρ ≠ 0 t = r 1− r2 n − 2 = 0.915 1− (0.915)2 5− 2 ≈ 3.928 t = r 1− r2 n − 2 = 0.92 1− (0.92)2 5− 2 ≈ 4.066 ˆy = b0 + b1 x b1 = xy− x y∑∑ n ∑ 2 x − 2 x∑( ) n ∑ = 11540− (72)(770) 5 1096− 2 72 5 ≈ 7.635 y = y∑ n = 770 5 =154, x = x∑ n = 72 5 =14.4 b0 = y − 1b x = 154 – (7.635)(14.4) = 44.056 ˆy = 44.056( ) + 7.635( ) x Page of6 7 1 3 1 1 1 1 1 1
  7. 7. Choose the one alternative that best completes the statement or answers the question. Consider the following median selling prices ($thousands) for homes in a community: (a) Use year 1 as a base year and construct a simple index number to show how the median selling price has increased. (b) Determine the actual percentage growth in the median selling price between the base year and year 4. (c) Determine the actual percentage growth in the median selling price between the base year and year 2. (d) Determine the actual percentage growth in the median selling price between year 2 and year 4. Solution Q7 : (a) (b) The median selling price in year 4 has an index of 115.33. This means that the median selling price in year 4 is 15.33% above the median selling price in year 1. (c) The median selling price in year 2 has an index of 103.33. This means that the median selling price in year 2 is 3.33% above the median selling price in year one. (d) To determine the actual percentage increase, the calculation is: ! * * * * * Q6 : Marks : 5 1 the χ2 (chi- square) statistic is distributed approximately as a chi- square only if the sample size is A. large (at least 30) B. too small (less than 30) C. between 20 and 30 2 A quantity under examination in an experiment as a possible cause of variation in the response variable is called a(n) A. Balanced Design B. Factor C. Levels 3 changes in time-series data that are unpredictable is called a(n) ___________ A. A random component B. A cyclical component C. A seasonal component 4 A company has a monthly time series that regularly shows sales being higher in the summer months. This is an example of which component? A. Cyclical B. Trend C. Seasonal 5 If you wish to test the claim that the mean of the population is 100, the appropriate null hypothesis is __________ A. x̅ = 100 B. µ ≥ 100 C. µ = 100 Q7 : Marks :5 Year Price ($) 1 300 2 310 3 330 4 346 5 362 Year Price ($) Index N 1 300 100.00 2 310 103.33 3 330 110.00 4 346 115.33 5 362 120.67 115.33−103.33 103.33 × 100 = 11.6% Page of7 7 1 1 1 2 1 1 1 1 1

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