A researcher wants to conduct a before/after study on 11 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be -2.40 with a sample standard deviation of 1.21. A 0.05 level of significance is selected. Assume that the differences are normally distributed in the population. The table t value for this test is ±1.8125

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A researcher wants to conduct a before/after study on 11 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero
while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from
the before scores, the average difference is computed to be -2.40 with a sample standard deviation of 1.21. A 0.05 level of significance is selected. Assume that the differences are normally
distributed in the population. The table t
value for this test is
+1.8125
+2.2622
+1.7959
+2.2281
+3.1693
Transcribed Image Text:A researcher wants to conduct a before/after study on 11 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be -2.40 with a sample standard deviation of 1.21. A 0.05 level of significance is selected. Assume that the differences are normally distributed in the population. The table t value for this test is +1.8125 +2.2622 +1.7959 +2.2281 +3.1693
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