A researcher wanted to determine if teaching approaches influence second graders’ ability to do columnar addition. To test this, the researcher randomly selected 75 second graders from a local school and assigned them to one of three teaching conditions: (1) rote memory, (2) manipulatives, or (3) computer. One group of 25 students were taught using rote memory, a second group of 25 students was taught using manipulatives, and a third group of 25 students was taught with a calculator. All students were given a test on columnar addition. The means and standard deviations for the three groups are shown below.Mean Number of Correct Answers for Three Groups of Second Grade Students on a Test of Columnar AdditionGroup n M SDRote Memory 25 4.00 2.24Manipulatives 25 9.00 3.08Calculator 25 17.00 2.74What hypothesis testing technique should the researcher use to analyze the data? Two-sample t-test Paired t-test z-test for the population mean ANOVA
Question
A researcher wanted to determine if teaching approaches influence second graders’ ability to do columnar addition. To test this, the researcher randomly selected 75 second graders from a local school and assigned them to one of three teaching conditions: (1) rote memory, (2) manipulatives, or (3) computer. One group of 25 students were taught using rote memory, a second group of 25 students was taught using manipulatives, and a third group of 25 students was taught with a calculator. All students were given a test on columnar addition. The means and standard deviations for the three groups are shown below.Mean Number of Correct Answers for Three Groups of Second Grade Students on a Test of Columnar AdditionGroup n M SDRote Memory 25 4.00 2.24Manipulatives 25 9.00 3.08Calculator 25 17.00 2.74What hypothesis testing technique should the researcher use to analyze the data? Two-sample t-test Paired t-test z-test for the population mean ANOVA
Solution
The researcher should use ANOVA (Analysis of Variance) to analyze the data. This is because there are more than two groups being compared. ANOVA is a statistical technique that is used to compare the means of more than two groups. The two-sample t-test and paired t-test are used when comparing the means of two groups, and the z-test for the population mean is used when comparing a sample mean to a population mean. In this case, the researcher is comparing the means of three groups (rote memory, manipulatives, and calculator), so ANOVA is the appropriate technique.
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Using four methods to teach ANOVA, do these four samples differ enough from each other to reject the null hypothesis that type of instruction has no effect on mean test performance?Method to teach ANOVA Mean SD Nmethod 1 (single teacher) 4.85 0.360 34method 2 (co-teachers) 4.61 0.715 31method 3 (computer) 4.61 0.688 36method 4 (lab) 4.38 0.793 32Since we are comparing more than 2 groups, we will use ANOVA to test whether the data provide evidence that test score is related to teaching method.One of the conditions that allows us to use ANOVA safely is that of equal (population) standard deviations. Can we assume that this condition is met in this case? Yes, since 0.793 − 0.360 < 2. No, since 0.793/0.360 > 2 No, since the four sample standard deviations are not all equal. No, since the population standard deviations are not given, so we cannot check this condition.
Using four methods to teach ANOVA, do these four samples differ enough from each other to reject the null hypothesis that type of instruction has no effect on mean test performance?Method to teach ANOVA Mean SD Nmethod 1 (single teacher) 4.85 0.360 34method 2 (co-teachers) 4.61 0.715 31method 3 (computer) 4.61 0.688 36method 4 (lab) 4.38 0.793 32Since we are comparing more than 2 groups, we will use ANOVA to test whether the data provide evidence that test score is related to teaching method.Let μ1, μ2, μ3, and μ4 be the mean test scores for teaching using methods 1, 2, 3, and 4 respectively. Which of the following are the appropriate hypotheses in this case?H0: µ1 = µ2 = µ3= µ4Ha: µ1, µ2, µ3, µ4 are not all equalH0: µ1 − µ2 < 0Ha: µ1 − µ2 = 0H0: µ1 = µ2 = µ3= µ4Ha: µ1 ≠ µ2 ≠ µ3≠ µ4H0: µ1 ≠ µ2 ≠ µ3 ≠ µ4Ha: µ1 = µ2 = µ3 = µ4
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