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before he got the new tool, Gary collected data for a random sample of 6 medium-sized rooms. He determined that the mean amount of time

before he got the new tool, Gary collected data for a random sample of 6 medium-sized rooms. He determined that the mean amount of time that it took him to paint each room was 4.2 hours with a standard deviation of 0.5 hours. For a random sample of 4 medium-sized rooms that he painted using the new tool, he found that it took him a mean of 3.9 hours to paint each room with a standard deviation of 0.7 hours. At the 0.10 level, can Gary conclude that his mean time for painting a medium-sized room without using the tool was greater than his mean time when using the tool? Assume that the population variances are equal. (Pooled yes) Ho:1 2 Ha: 1 > 2 = 0.797 Test Statistic = (1 2) (1 2) ( 2 1 1 + 2 2 2) = (4.2 3.9) (0) ( .52 6 + .72 4 ) Critical Value: invT(.10): 1.397 p-value: 0.224 on calculator p-value 0.224 > 0.10 Fail to Reject Conclusion: At the 0.10 level of significance, there is not sufficient evidence to say that there is a reduction in Gary's mean time to paint a room using the new tool. A SAT prep course claims to increase student scores by more than 60 points, on average. To test this claim, 9 students who have previously taken the SAT are randomly chosen to take the prep course. Their SAT scores before and after completing the prep course are listed in the following table. Test the claim at the 0.01 level of significance. The original answer was incorrect. The there was a z value when there was not a population standard deviation. The one who did this problem may have gotten their letters mixed up. The Critical Value is 1.397, because when you do your degrees of freedom and you have two samples, you need to add the number of samples together, then subtract two from it.

SAT Scores Before Prep Course 1010 980 1170 1200 1040 1280 1450 1470 1500 After Prep Course 1100 1260 1190 1280 1170 1370 1440 1500 1520 Difference 90 280 20 80 130 90 -10 30 20 Ho: 60 Ha: > 60 = .728 Test Statistic = ( ) = 81.88 60 (86.95 9 ) Critical Value: invT(.01, df: 9-1 = 8 ): 2.896 p-value: 0.2436 on Calculator 0.2436 > 0.01 Fail to Reject Conclusion: At the 0.01 level of significance, there is not sufficient evidence at the 0.01 level of significance to support the claim that students SAT scores increase by a mean of more than 60 points after completing the SAT prep course

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