For an experiment comparing more than two treatment conditions, youshould use an

Question

For an experiment comparing more than two treatment conditions, youshould use analysis of variance rather than separate t testsbecause
a) you are less likely to make a mistake in thecomputations of ANOVA b) a test based on variances is more sensitive than atest based on means c) ANOVA has a less risk of a Type 1 error d) ANOVA has a less risk of a Type II error
Please help: I am not sure of what is a Type 1 error orType II error. Thanks
a) you are less likely to make a mistake in thecomputations of ANOVA b) a test based on variances is more sensitive than atest based on means c) ANOVA has a less risk of a Type 1 error d) ANOVA has a less risk of a Type II error
Please help: I am not sure of what is a Type 1 error orType II error. Thanks a) you are less likely to make a mistake in thecomputations of ANOVA b) a test based on variances is more sensitive than atest based on means c) ANOVA has a less risk of a Type 1 error d) ANOVA has a less risk of a Type II error
Please help: I am not sure of what is a Type 1 error orType II error. Thanks

Explanation / Answer

Explanation: It is tempting to test the null hypothesisH0:1=2=3by comparing the population means two at a time using the t-tests.If we proceed this way , we would need to test three differenthypotheses: H0:1=2 andH0:1=3 andH0:2=3 Each hypothesis would have a probabiliy of Type I error(rejecting the null hypothesis when it is true) of . If weused an =0.05 level of signficance, each hypothesis wouldhave a 95% probability of rejecting the null hypothesis when thealternative hypotheis is true (i.e. a 95% probabilty of making acorrect decision). The probabiltiy that all three hypothesescorrectly reject the null hypothesis is 0.953=0.86(assuming the tests are independent). There is a1-0.953=1-0.86=0.14, or 14%, probabillty that at leastone hypothesis will lead to an incorrect rejection ofH0. A 14% probabilty of a Type I error is much higherthan the desired 5% probability. As the number of populations thatare to be compared increases, the probability of making a Type Ierror using multiple t-tests for a given value of increases. To address this problem, sir Ronald A.Fisher introduced themethod of analysis of variance Each hypothesis would have a probabiliy of Type I error(rejecting the null hypothesis when it is true) of . If weused an =0.05 level of signficance, each hypothesis wouldhave a 95% probability of rejecting the null hypothesis when thealternative hypotheis is true (i.e. a 95% probabilty of making acorrect decision). The probabiltiy that all three hypothesescorrectly reject the null hypothesis is 0.953=0.86(assuming the tests are independent). There is a1-0.953=1-0.86=0.14, or 14%, probabillty that at leastone hypothesis will lead to an incorrect rejection ofH0. A 14% probabilty of a Type I error is much higherthan the desired 5% probability. As the number of populations thatare to be compared increases, the probability of making a Type Ierror using multiple t-tests for a given value of increases. To address this problem, sir Ronald A.Fisher introduced themethod of analysis of variance

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