Quiz 2 for Business Statistics II | STAT 202, Quizzes of Business Statistics

Download Quiz 2 for Business Statistics II | STAT 202 and more Quizzes Business Statistics in PDF only on Docsity! STAT 202 Quiz #2 Fall 2004 1. Consider testing H0: µ ≥ 40 vs. H1: µ < 40 where µ represents the mean of a normally distributed population having known standard deviation σ = 10. The test will be based on a sample of size n = 25, with significance level α = 0.01. In terms of the sample mean X , what is the decision rule (rejection region) for this test? Your answer should look like ‘reject H0 if X > 84.27’ or ‘reject H0 if X < 21.33’, etc. On the z-scale, you’d reject H0 if z < -2.33, so you’d reject if the sample result is more than 2.33 standard deviations below the mean. On the X scale, that translates to 40 – 2.33(10/sqrt(25)) = 35.34 (your answer may vary a little due to rounding). So we reject H0 if X < 35.34. 2. Consider testing H0: µ ≤ 60 vs. H1: µ > 60 where µ represents the mean of a normally distributed population having known standard deviation σ = 5. The test will be based on a sample of size n = 30, with significance level α = 0.05. The decision rule for this test is to reject H0 if X > 61.50. Calculate the power of this test when H1 is true and µ is actually equal to 62. This is a normal curve probability problem: you want the probability that X > 61.50, where X has a normal distribution with mean equal to 62 and standard deviation 5/sqrt(30) = 0.9129. Draw the sketch, calculate z = -0.55. Solution is 0.5 + 0.2088 = 0.7088. A B C D E F G H I 0.1436 0.2912 0.4602 0.5398 0.6144 0.7088 0.8564 0.9525 0.9984 3. Consider testing H0: π ≥ 0.20 vs. H1: π < 0.20, based on a sample of size n = 240. Suppose that you're told that, when H1 is true and the actual population proportion is π = 0.16, the power of the test is 0.6266. This tells us that: (a) The probability of making a Type II error is 0.6266. (b) The probability of rejecting H0 when in fact π = 0.16 is 0.6266. (c) The probability of not rejecting H0 when in fact π = 0.16 is 0.6266. (d) The probability of rejecting H0 when in fact π = 0.20 is 0.6266. (e) The probability of not rejecting H0 when in in fact π = 0.20 is 0.6266. 4. Consider testing H0: µ ≥ 25 vs. H1: µ < 25 where µ is the mean of a normally distributed population. Which sketch shows the shape of the power function for this test? It’s (a). Power must increase as you move deeper into the “H1 region”, which in this case means as move further from 25 to the left. (a) (b) (c) 0 0.2 0.4 0.6 0.8 1 22 23 24 25 26 27 28 µ P {R ej ec t H 0} 0 0.2 0.4 0.6 0.8 1 22 23 24 25 26 27 28 µ P {R ej ec t H 0} 0 0.2 0.4 0.6 0.8 1 22 23 24 25 26 27 28 µ P {R ej ec t H 0} Two different machines are used to automatically fill “16 oz.” boxes of Special Z cereal (the actual fill weights vary, as the machines aren’t perfect). When operating properly, both machines should have the same variation in fill weights. To check this we test H0: σ12 = σ22 vs. H1: σ12 ≠ σ22. The test will use the following data, taken from a random sample of filled boxes from each machine’s output. Fill weights are assumed to be normally distributed. n1 = 8 033.161 =x oz. s1 = 0.47 oz. n2 = 10 219.162 =x oz. s2 = 0.93 oz. 5. The calculated value of the appropriate test statistic is: It’s the ratio of the larger to the smaller sample variance: (.932/.472) = 3.92. (a) 0.2554 (b) 0.4177 (c) 0.5054 (d) 0.8321 (e) 1.9787 (f) 2.4476 (g) 3.9153 (h) 4.2388 6. At the α = 0.05 level of significance, this test would reject the null hypothesis if the calculated test statistic is greater than: It’s the upper 0.025 tail of the F-distribution with 10 – 1 = 9 DF in the numerator and 8 – 1 = 7 DF in the denominator. That’s 4.82 from the table. (a) 2.45 (b) 3.29 (c) 3.68 (d) 4.20 (e) 4.82 (f) 5.14 (g) 5.57 (h) 6.63 It is also important that the average fill weight is the same for both machines, so we wish to test: H0: µ1 = µ2 vs. H0: µ1 ≠ µ2 . Suppose that the specified level of significance is α = 0.05, and that we can assume that the variability in fill weights is the same for both machines. (continued next page) Solutions: 1. Reject H0 if ___________________________ 2. _______ 3. _______ 4. _______ 5. _______ 6. _______ 7. _______ 8. _______ 9. _______ 10. _______ 11. _______ 12. ______ Formulas: Hypothesis testing: n XX z X σ µ σ µ 00 −= − = ; n zX σµ += 0 Two-Sample Tests: ( ) ( ) 2 2 2 1 2 1 21 2 2 2 1 2 1 21 21 21 2 22 2 112 p 21 2 21 or 2 DF 2 11s where 11 nn xxz n s n s xxz nn nn snsn nn s xxt p σσ + − = + − = −+= −+ −+− =       + − = 21 2211 21 21 where 11)1( 1 :freedom of Degrees / nn pnpnp nn pp ppz n ns dt d + + =       +− − = −= F = maximum{ 2 1 2 2 2 2 2 1 , s s s s } One-Way Analysis of Variance: )/( )1/( tNSSE tSSTR MSE MSTRF − − == Degrees of freedom: (t-1, N – t) (Formulas for randomized block designs are not given: we’ll just work with computer output.) Standard deviation: ( ) 11 222 − − = − − = ∑∑ n dnd n dds iid