To evaluate the effect of a treatment, a sample of n = 8 is obtained from a population with a mean of μ = 50, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 55.Assuming that the sample variance is s^2 = 32 , use a two-tailed hypothesis test with α = .05 to determine whether the treatment effect is significant and compute both Cohen’s d and r^2 to measure effect size.Assuming that the sample variance is s^2 = 72 , repeat the test and compute both measures of effect size.Comparing your answers for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test and measures of effect size?

Name: To evaluate the effect of a treatment, a sample of n = 8 is obtained
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Description: To evaluate the effect of a treatment, a sample of n = 8 is obtained from a population with a mean of μ = 50, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 55.Assuming that th

1. The null and alternative hypothesis are given by : Ho : μ=50H1 : μ≠50 The test statistics is given by Students' t-distribution : t=x-μs2/n~tn-1 t=55-5032/8=2.5 The degrees of freedom are (8-1)=7 The two-tailed critical t-value at α=0.05, df=7 is ±2.365 The decision rule states : Null hypothesis is rejected when calculated test statistics is greater than the tabulated t-value and is accepted other wise. Since the calculate value 2.5 is greater than tabulated value 2.365, the null hypothesis is rejected concluding that the treatment does not have any significant effect. Cohen's d formula is given by : d=x-μs=55-5032=0.88 The Cohen's d-value indicates a large effect. The formula for R2 is given by : R2=t2t2+df=2.522.52+7=0.472 2. For sample variance s2=72, the test statistics and the measure of effect sizes are : t=55-5072/8=1.67 For the same critical value, the null hypothesis this time is accepted because the calculated test statistics 1.76 is less than the tabulated value 2.365, concluding that the population mean is not equal to 50. Cohen's d and r2 in this case are : d=55-5072=0.589 There is a medium effect. r2=1.6721.672+7=0.285 3. The increase in the variability of the scores decreases the test statistics, the Cohen's d and the R2

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To evaluate the effect of a treatment, a sample of n = 8 is obtained from a population with a mean of μ = 50, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 55.

Assuming that the sample variance is s^2 = 32 , use a two-tailed hypothesis test with α = .05 to determine whether the treatment effect is significant and compute both Cohen’s d and r^2 to measure effect size.
Assuming that the sample variance is s^2 = 72 , repeat the test and compute both measures of effect size.
Comparing your answers for parts a and b, how does the variability of the scores in the sample influence the outcome of a hypothesis test and measures of effect size?

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