KING FAHD UNIVERSITY OF PETROLEUM & MINERALS DEPARTMENT OF MATHEMATICS & STATISTICS DHAHRAN, SAUDI ARABIA

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1 SECTION # SERIAL # ID# NAME: KING FAHD UNIVERSITY OF PETROLEUM & MINERALS DEPARTMENT OF MATHEMATICS & STATISTICS DHAHRAN, SAUDI ARABIA STAT 39: PROBABILITY & STATISTICS FOR ENGINEERS & SCIENTISTS Final Examination for Semester 083 on Wednesday, nd September, between 0900 pm and 30 pm at Exhibition Center Please circle the name of your instructor; Write section number, serial number, ID number (with all zeroes) and name (in Capital or block or upper case letters) on the cover page and on any addendum. Instructor Section Time of lecture Esam AlSawi 3 Anwar Joarder / am to 00 am 030 am to 30 am Put the choices of Question 4 in the following table: A B I I A Q S/M Question Marks Marks Obtained 9 Strengths/Weakness of the Student Total 80

2 SECTION # SERIAL # ID# NAME:. (8+7+4=9 Marks) A production filling operation has a historical standard deviation of 6 ounces. When in perfect adjustment, the mean filling weight for the production process is 50 ounces. A quality control inspector periodically selects at random 36 containers and uses the sample mean filling weight to see if the process is in perfect adjustment a. [++++= 8 Marks]Test the hypothesis at the 5% level of significance if the sample mean filling weight is 48.6 ounces. Write clearly the hypotheses, the test statistic, Rejection and Acceptance Regions for the null hypothesis, the sample value of the test statistic, the decision. Population: Let X be the filling weight and X N ( µσ, ) = N ( µ,6 ). Objective: Testing at 5% level of significance H = 50 0 against H 50. Method: Since σ = 6, an exact z -test would be appropriate. The test statistics is Z X µ =. σ / n Rejection Region for the null hypothesis: { z : z <.96, or, z >.96}. x µ Decision: Since the sample produces z = = =.4 which is not σ / n 6 / 36 among the largest (or smallest).5% of the possible z -values, hence we accept the null hypothesis. b. [3++ =7 Marks] Derive the p-value of the test and make a decide on part (a). Also report two values of significance level for which one rejects the null hypothesis and two values of significance level for which one accepts the null hypothesis. The probability value is given by p-value = P( Z <.4) = (0.0808) = Hence one accepts the null hypothesis at α < 0.66, say at 0.0, 0.05, 0.04, 0.03 etc; one rejects if α 0.66 (unsual)! c. [3+ = 4 Marks] Develop a 95% confidence interval and use it to test the hypothesis in part (a). 95% CI for the expected filling weight is given by x z n, 0.05 σ / = / 36 = or, < µ < H = 50 0 is inside the above CI, we accept H = 50 0.

3 SECTION # SERIAL # ID# NAME: 3. [++++= 8 Marks] Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of n = 0 gears from the first supplier results in x = 90 and s =, while another random sample of n = 6 gears from the second supplier results in x = 3 and s =. Test at % level of significance the claim that the mean impact strength of gears from Supplier is at least 5 foot-pounds higher than that of Supplier. Write clearly the hypotheses, the test statistic, Rejection and Acceptance Regions for the null hypothesis, the sample value of the test statistic, the decision. Population: Let X be the impact strengths of the gears from the first supplier and X be the impact strengths of the gears from the second supplier X i N ( µ i, σ ), i =,. Objective: Testing at 5% level of significance H 0 µ 5, against H µ < 5 i.e., H 0 µ 5 against H µ > 5. Method: Since σ is unknown, an exact t -test would be appropriate. The test statistics is T ( X X ) ( µ µ ) =. ( S / n ) + ( S / n ) Rejection Region for the null hypothesis: { t : t >.49}. 9( ) + 5( ) Decision: Since the sample produces s = = 356.5, ( x x ) ( µ µ ) (90 3) ( 5) t = = = which is not among the ( s / n) + ( s / n) ( s / + 0 ( s )/ ) 5 largest % of the possible z -values, hence we accept the null hypothesis.

4 SECTION # SERIAL # ID# NAME: 4 3. [5 Marks] An engineer wanted to analyze the relationship between the speed of a car (x) and its gas mileage (y). As an experiment a car is operated at several different speeds and for each speed the gas mileage is measured. These data are shown below. Speed Gas Mileage a. [6++=8 Marks] Derive an equation of the least squares regression line for the above sample. Draw the regression line. Check if the line passes through the average point ( x, y ). The sample total and sum of squares for speed ( x ) are given by 350 and 900 respectively so that centered sum of squares is s xx = 600. The sample total and sum of squares for gas mileage ( y ) are given by 3 and 7833 respectively so that so that centered sum of squares s yy = 0. The sum of products between speed and gas mileage is 0985 so that the centered sum of products is s xy = 565. Hence an estimate of the slope is ˆ 565 β = = , and that of slope is ˆ β 0 = Hence the regression line is 600 ˆ( µ x) = x. The average point (50, 33) passes through the regression line.

5 SECTION # SERIAL # ID# NAME: 5 b. [+ = Marks] Estimate the expected gas mileage for a car having speed 5 mph, by using the regression line. Explain why we can or cannot estimate the expected gas mileage for a car having speed 5 mph by y only. ˆ(5) µ = (5) = The sample mean y = 33 will underestimate the expected gas mileage for a car having speed 5 mph c. [ Mark] Estimate the error in estimating the expected gas mileage for a car having speed 5 mph. y ˆ(5) µ = =.885. If ˆ(5) µ is estimated by y = 33, the error will be = 7. d. [ Mark] Estimate the expected gas mileage of a car traveling at 55 mph by a point. ˆ(55) µ = (55) = e. [ Marks] Estimate the variance of the gas mileages. Since CSS = 0, SSR = ˆ βsxy = , SSE = , the variance σ is estimated by MSE SSE /( n ) = =. f. [++++ = 8 Marks] Test the hypothesis at 5% level of significance that the higher the speed the lesser the gas mileage. Write clearly the hypotheses, the test statistic, Rejection and Acceptance Regions for the null hypothesis, the sample value of the test statistic, the decision. Population: Let for a given speed x of a car, the gas mileage Y ~ N ( β0 + βx, σ ) Objective: Testing at 5% level of significance H0 : β 0 against H: β < 0. Method: Since σ is unknown, an exact t -test would be appropriate. The test statistics is T ˆ β 0 =. MSE / S xx Rejection Region for the null hypothesis: { t : t <.05}. Decision: Since the sample produces, ˆ β t = = = mse / s xx /600 which is among the smallest 5% of the possible t -values, hence we reject the null hypothesis.

6 SECTION # SERIAL # ID# NAME: 6 g. [3 Marks] Predict by an interval the gas mileage of a car traveling at 55 mph by using 95% confidence. ( x x) ˆ( µ x ) tα / + + MSE n sxx (55 50) = = MSE Hence the gas mileage is predicted to be in between and Choose the best choice (not the partially true one) in Question 4 and record on to the first page. Note that NOTPC means None Of The Previous Choices. Each question 4.x carries 4 marks. 4. The number of arrivals at a local gas station between 3:00 and 5:00 P.M. has a Poisson distribution with a mean of. Find the probability that the number of arrivals between 4:00 and 5:00 P.M. is exactly two. C A. 6 8e I R. 7e O. NOTPC ( t ) x λ λ t PX ( = x) = e, x! ( 0.5) (0.5) 6 PX ( = ) = e = 8e ! 4. The time it takes a technician to fix a computer problem is exponentially distributed with a mean of 5 minutes. What is the probability that it will take a technician between 0 to 5 minutes to fix a computer problem? D. e /3 /(5) U B. /3 ( e ) / e A I. NOTPC P(0 < X< 5) = PX ( > 0) PX ( > 5) = e e = e e = /3 ( e ) / e /5 5/5 /3 4.3 In order to ensure efficient usage of a server, it is necessary to estimate the mean number of concurrent users. According to records, the number of concurrent users follow a normal distribution with a mean of 75 and standard deviation 6. The number of users are checked at 50 different times. Solve the equation PX ( < k) = 0.9 for k. S Y..8 R..8 I A. NOTPC By standardizing, the given equation can be written as

7 SECTION # SERIAL # ID# NAME: 7 X 75 k 75 P < = / 50 6 / 50 k 75 which yields =.8 which yields 6 / 50 k = µ + z σ n = = ( 0.88) / / The power of supercomputers derives from the idea of parallel processing. Engineers at Cray Research are interested in determining whether one of two parallel processing designs produces faster average computing time, or whether the two designs are equally fast. The following are the results, in seconds, of independent random computation times using the two designs. Design Design Write the hypothesis to check the evidence that one parallel processing design allows for faster average computation than the other? I. H a < µ, or, µ > µ, AH. a µ, The symbols have usual meaning. I. H a < µ, or, µ > µ. < L H a µ µ. : >, Y. NOTPC 4.5 A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and 8 of these helmets some damage was observed. Using the point estimate of p obtained from the preliminary sample of 50 helmets, how many helmets must be tested to be 95% confident that the error in estimating the true value of p is less than 0.0? S. 40 P. 69 A. 3 I. 559 N. NOTPC z α /.96 n pq ˆˆ = = (0.36)(0.64) =.766 e There are 9 devices in a box. Five of them are sound and four are faulty. One takes four devices in succession at random without replacement. What is the probability that the first two are faulty but the last two are sound? Q. 5/63 A. 30/63 T. 5/63 I. 0/63 F. NOTPC P( FFS 3S4) = = =

8 SECTION # SERIAL # ID# NAME: There are N devices in a box. A total of K of them are faulty and N K are sound. One takes three devices in succession at random without replacement. The probability that first two are faulty but the third one is not faulty is / 60. What is the probability that in a sample of 3 devices, two are faulty and one is sound? S. 3/5 M. 6/0 I. 9/0 L. 6/56 E. NOTPC 3 PX ( = ) = ( / 60) = 3/ 5. Alternatively, K + 0 ( K ) ( N K) P( FFS 3) =, K + ( N K) ( K ) + ( N K) ( K + ( N ) K) By the given condition, we have preparing the following table: K( K )( N K) = N ( N )( N ) 60 which can be solved by K N K N P( FFS 3) approx approx Notice that there are two solutions K = 3, N K =, or, K = 4, N K =. Hence, we have K N K N PX= ( ) = = 3 3 5