PUAD660 - Quantitative Methods Unit 5 Practice Problems Answer Key https://books.google.com/books? id=8VU8AwAAQBAJ&pg=PA222&lpg=PA222&dq=Random+samples+of+men+and+women+ have+been+given+a+scale+that+measures+their+support+for+gun+control.&source=bl&ots=64 Ta1izXIT&sig=fw7wZGqijOzEKU3AuOBlceDTpfY&hl=en&sa=X&ved=0ahUKEwi51667_ZT TAhXD0iYKHcNyDH8Q6AEINjAF#v=onepage&q=Random%20samples%20of%20men %20and%20women%20have%20been%20given%20a%20scale%20that%20measures%20their %20support%20for%20gun%20control.&f=false Part 1. Mixed Multiple Choice Questions 1. B 2. D 3. B 4. D 5. B 6. C 7. D 8. B Part 2. Two sample hypothesis test Step 1. Making Assumptions 1. Samples 1 and 2 are random and independent. 2. Our variable is interval ratio (Score) 3. The sampling Z distribution is normal (our sample is large enough (N>120 for both samples together)) Step 2. Stating the Hypotheses Research Hypothesis: There is a difference between football and basketball players in the population. Null Hypothesis: There is no difference between football and basketball players in the population. Any difference observed between samples occurred by pure chance. Step 3. Drawing the Sampling Distribution and establishing a critical region. PUAD660 - Quantitative Methods Zobtained=1.7 Step 4. Computing the test statistic Basketball X1 = 460 s1 = 92 N1 = 102 Football X2 = 442 s2 = 57 N2 = 117 PUAD660 - Quantitative Methods X X X X s1 2 s 22 N 1 1 N 2 1 (92) 2 (57) 2 102 1 117 1 X X 8464 3249 101 116 X X 83.80 28.01 X X 111.81 X X 10.57 Z(obtained) ( X 1 X 2 ) (460 442) 18 1.70 X X 10.57 10.57 Step 5. Placing Z obtained on the distribution and stating the conclusion. Conclusion: Z obtained is not in the critical region (it is not higher than the Z critical). We fail to reject the null hypothesis with 95% confidence. It appears that there is no difference between football and basketball players in the population. Part 3. Two sample hypothesis test Step 1. Making Assumptions 1. Samples 1 and 2 are random and independent. 2. Our variable is a long ordinal scale (rating) 3. The sampling Z distribution is normal (our sample is large enough (N>120 for both samples together)) Step 2. Stating the Hypotheses Research Hypothesis: There is a difference between AA and Reg employees in terms of efficiency. Null Hypothesis: There is no difference between AA and Reg employees in terms of efficiency. Step 3. Drawing the Sampling Distribution and establishing a critical region. PUAD660 - Quantitative Methods -0.83 Step 4. Computing the test statistic Sigma = 2 2 S1 S2 8.41 4 + = + = 0.13=.36 N 11 N 21 96 99 Z(obtained) = 15.215.5 =0.83 0.36 Step 5. Placing Z obtained on the distribution and stating the conclusion. Conclusion: Z obtained is not in the critical region. We fail to reject the null hypothesis with 95% confidence. It appears that there is no difference between AA and Reg employees in terms of efficiency. Part 4. Two sample one tailed test Step 1. Making Assumptions PUAD660 - Quantitative Methods 1. Samples 1 and 2 are random and independent. 2. Our variable is nominal (test of proportion) 3. The sampling Z distribution is normal (our sample is large enough (N>120 for both samples together)) Step 2. Stating the Hypotheses Research Hypothesis: The program resulted in a significantly lower divorce rate. Null Hypothesis: There is no difference in divorce rates across programs. Step 3. Drawing the Sampling Distribution and establishing a critical region. -0.76 -1.65 Conducting a one tailed test, with a negative tail (lower divorce rate). Step 4. Computing the test statistic Special Ps1 = .53 N1 = 78 Pu Regular Ps2 = .59 N2 = 82 N 1 Ps1 N 2 Ps 2 (78)(.53) (82)(.59) 41.34 48.38 0.56 N1 N 2 78 82 160 PUAD660 - Quantitative Methods p p Pu (1 Pu ) Z (obtained) = N1 N 2 78 82 160 (.56)(.44) .2464 (.4964)(.1582) .079 N 1N 2 (78)(82) (6396) ( Ps1 Ps 2) .53 .59 .06 0.76 pp .079 .079 Step 5. Placing Z obtained on the distribution and stating the conclusion. Conclusion: Z obtained is not in the critical region. We fail to reject the null hypothesis with 95% confidence. It appears that there is no difference between the experimental and traditional programs in terms of divorce rates