Math 628 In-class Exam 2 04/03/2013

Transcription

1 Math 628 In-class Exam 2 04/03/2013 Name: KU ID: Note: Show ALL work clearly in the space provided. In order to receive full credit on a problem, solution methods must be complete, logical and understandable. 1. (16pts) In a mechanical testing lab, Plexiglass strips are stretched to failure. Let X be the change in length in mm before breaking. Assume that the distribution of X is N(µ, σ 2 ). We shall test the null hypothesis H 0 : µ = 5.70 against the alternative hypothesis H 1 : µ 5.70, using n = 8 observations of X. (a) Define the test statistic and a critical region that has a significance level of α = (b) A random sample of eight observations of X yielded the following data: Calculate the value of the test statistic and state your conclusion clearly. (c) Give the approximating value or bounds for the p-value of this test. Solution: (a) The test statistic is T = X µ S 2 n and the critical region is { T t0.025 (7) = }. (b) From the set of data, x = 5.87 and s 2 = The value of the test statistic is t = = We reject H (c) Since t (7) = < < = t 0.01 (7), the p-value 2P (T > 2.435) is between 0.02 and (16pts) Let p be the proportion of college students who favor a new policy for alcohol consumption on campus. (a) How large a sample is required to estimate p so that the maximum error of the estimate of p is 0.02 with 95% confidence.

2 (b) Determine the sample size in part (a) when the size of the student body is Solution: (a) The desired sample size is n = z = (b) The required sample 4(0.02) 2 size is n = m 1+ m 1 N = (0.02) (0.02) = (16pts) According to a population census in 1986, the percentage of males who are 18 or 19 years old and are married was 3.7%. We shall test whether this percentage increased from 1986 to (a) Define the null and alternative hypotheses. (b) Define a critical region that has an approximate significance level of α = (c) If y = 20 out of a random sample of n = 300 males, each 18 or 19 years old, were married. What is your conclusion? What is the p-value of the test? Solution: (a) H 0 : p = 3.7% and H 1 : p > 3.7%. (b) The critical region is { z = y n ( ) (c) The computed value of the test statistic is z = n z 0.01 = } ( ) 300 = > z We reject H 0 and say that the percentage increased from 1986 to The p-value is P (Z 2.722) = (20pts) Two different saws are used to cut columns for a gazebo that has a diameter of 16 feet. Suspecting that one saw is cutting columns shorter than the other saw, the company is interested in comparing the two saws. Let the lengths of the columns cut by the two saws be observations of random variables X and Y that are N(µ X, σx 2 ) and N(µ Y, σy 2 ), respectively. We shall use the following observations of X and Y : Assume that α = X Y (a) Test H 0 : σ 2 X = σ2 Y against H 1 : σ 2 X σ2 Y. (b) Use the result in part (a) to test H 0 : µ X = µ Y against H 1 : µ X < µ Y. Solution: (a) From the set of data, x = 8.05, y = , s 2 X = and s 2 Y = Since s2 X = < 4.99 = F s (7, 7), we do not reject H 0. Y

3 (b) From part (a), we can assume σx 2 = σ2 Y. We use the t test. The test statistic is T = X Y X Y =. (n 1)SX 2 +(m 1)S2 Y ( ) SX 2 +S2 Y n+m 2 n m n So the value of the test statistic is t = We do not reject H = > = t 0.05 (14). 5. (16pts) Montgomery examined the strengths of a synthetic fiber that may be affected by the percentage of cotton in the fiber. Three levels of this percentage are considered, with four observations taken at each level. Percentage of Cotton Tensile Strength in Pounds per Square Inch Use the F test, with α = 0.05, to see if there are differences in the breaking strengths due to the percentages of cotton used. Solution: We first compute the means Now we compute the sum of squares: and X i Grand mean X 14 SS(T ) = 4[(10 14) 2 + ( ) 2 + ( ) 2 ] = SS(E) = 4 4 (X ij X i ) 2 = 89.5 i=1 j=1 Therefore, F = SS(T ) m 1 SS(E) = = > 4.26 = F 0.05 (2, 9). We reject H 0 and say that n m 12 3 there are differences in the breaking strengths due to the percentages of cotton used.

4 6. (16pts) In order to test whether four brands of gasoline give equal performance in terms of mileage, each of three cars was driven with each of the four brands of gasoline. The number of miles per gallon in each combination is recorded in the following table: Brand 1 Brand 2 Brand 3 Brand 4 Car Car Car Assume that the significance level α = Solution: We first compute the means Brand 1 Brand 2 Brand 3 Brand 4 X i Car Car Car X j Now we compute the sum of squares SS(A) = 4[( ) 2 + ( ) 2 + ( ) 2 ] = 7.26 SS(B) = 3[( ) 2 + ( ) 2 + ( ) 2 + ( ) 2 ] = and SS(E) = = We need to test whether four brands of gasoline give equal performance in terms of mileage. That is, we shall test the hypothesis H B : β 1 = β 2 = β 3 = β 4 = 0 against all alternatives. The calculated value of the test statistic is F B = < 4.76 = F 0.05 (3, 6). We do not reject H 0. SS(B) b 1 SS(E) (a 1)(b 1) = (3 1)(4 1) =

5 α z α z r t 0.25 (r) t 0.10 (r) t 0.05 (r) t (r) t 0.01 (r) F 0.05 (r 1, r 2 ) r 1 = r 2 = F (r 1, r 2 ) r 1 = r 2 =