Please look at the solution below: I have two doubts: a. critical value: why is it -ve?? should it not be 1.28 b. the z-statistic calculated (24.2 ¨C 25) / (1.5 / ¡Ì 100) = -5.333 should it not be (24.2 ¨C 25) / (.015 / ¡Ì 100) = -533.3 Please help!!! -------------------------------------------------------- Maria Huffman is the Vice President of Human Resources for a large regional car rental company. Last year, she hired Graham Brickley as Manager of Employee Retention. Part of the compensation package was the chance to earn one of the following two bonuses: if Brickley can reduce turnover to less than 30 percent, he will receive a 25 percent bonus. If he can reduce turnover to less than 25 percent, he will receive a 50 percent bonus (using a significance level of 10 percent). The population of turnover rates is normally distributed. The population standard deviation of turnover rates is 1.5 percent. A recent sample of 100 branch offices resulted in an average turnover rate of 24.2 percent. Which of the following statements is TRUE? A) Brickley should not receive either bonus. B) For the 25% bonus level, the test statistic is -10.66. C) For the 50% bonus level, the critical value is -1.65 and Huffman should give Brickley a 50% bonus. D) For the 50% bonus level, the test statistic is -5.33 and Huffman should give Brickley a 50% bonus. Your answer: A was incorrect. The correct answer was D) For the 50% bonus level, the test statistic is -5.33 and Huffman should give Brickley a 50% bonus. Using the process of Hypothesis testing: Step 1: State the Hypothesis. For 25% bonus level - Ho: m ¡Ý 30% Ha: m < 30%; For 50% bonus level - Ho: m ¡Ý 25% Ha: m < 25% . Step 2: Select Appropriate Test Statistic. Here, we have a normally distributed population with a known variance (standard deviation is the square root of the variance) and a large sample size (greater than 30.) Thus, we will use the Z-statistic. Step 3: Specify the Level of Significance. a = 0.10. Step 4: State the Decision Rule. This is a one-tailed test. The critical value for this question will be the Z-statistic that corresponds to an a of 0.10, or an area to the left of the mean of 40% (with 50% to the right of the mean). Using the z-table (normal table), we determine that the appropriate critical value = -1.28 (Remember that we highly recommend that you have the ¡°common¡± Z-statistics memorized!) Thus, we will reject the null hypothesis if the calculated test statistic is less than -1.28. Step 5: Calculate sample (test) statistics - Z (for 50% bonus) = (24.2 ¨C 25) / (1.5 / ¡Ì 100) = -5.333. Z (for 25% bonus) = (24.2 ¨C 30) / (1.5 / ¡Ì 100) = -38.67 . Step 6: Make a decision. Reject the null hypothesis for both the 25% and 50% bonus level because the test statistic is less than the critical value. Thus, Huffman should give Soberg a 50% bonus. The other statements are false. The critical value of ¨C1.28 is based on the significance level, and is thus the same for both the 50% and 25% bonus levels.
The question is whether, at the 10% significant level, we can reject the null hypothesis that the population mean is greater than or equal to 25, when we have a sample mean of 24.2 and a population standard deviation of 1.5. The answer is yes: the null hypothesis should be rejected and Brickley should receive his 50% bonus. The standard error of the sample mean here is 1.5/sqrt(100) = .15. The test statistic is therefore 24.2-25 divided by .15, or -5.33. The critical test statistic is -1.28, since this is a one-tailed test. Since -5.33 is less than -1.28, we reject the null hypothesis at the 10% significance level and award Brinkley his bonus.
The null hypothesis is H0: Mu0 >= 25% and alternative hypothesis is Mu0<25% The rejection rejection for which the alternative hypothesis claim is tested is on the left tail of the normal distribution. Since it is one tail test, all of Alpha is present in the left tail. The t-critical value would there fore be -ve, and the null hypothesis cannot be rejected if the t-calc value is greater(on the right side to) than t-critical. In the calcuation of t-calc , (24.2 - 25)/(1.5/ sqrt(100)), need to be consistent in both numerator and denominator as each variable is in percentage term.
I find this step 6 a bit unusual. Isn’t it suppose to say “Reject the null hypothesis for both the 25% and 50% bonus level because the test statistic is MORE than the critical value”? Any thoughts?
Why is the critical test statistic in this problem -1.28? Where do you find that number? Do you just have to memorize it?