Question

1. solve the following problems.

a. a repeated - measures study with a sample of ( n = 6 ) participants produces a mean difference of ( m _ { d } = 4 ) with ( s s = 30 ). use a two - tailed hypothesis test with ( alpha = 0.05 ) to determine whether this sample provides evidence of a significant treatment effect.

Explanation:

Step1: Calculate the variance of the difference scores

The formula for the variance of the difference scores \(s_{D}^{2}=\frac{SS}{n - 1}\). Given \(SS = 30\) and \(n=6\), then \(s_{D}^{2}=\frac{30}{6 - 1}=\frac{30}{5}=6\)

Step2: Calculate the estimated standard error

The formula for the estimated standard error \(s_{M_{D}}=\sqrt{\frac{s_{D}^{2}}{n}}\). Substituting \(s_{D}^{2} = 6\) and \(n = 6\), we get \(s_{M_{D}}=\sqrt{\frac{6}{6}}=\sqrt{1}=1\)

Step3: Calculate the t - statistic

The formula for the t - statistic in a repeated - measures design is \(t=\frac{M_{D}}{s_{M_{D}}}\). Given \(M_{D}=4\) and \(s_{M_{D}} = 1\), then \(t=\frac{4}{1}=4\)

Step4: Determine the critical t - values

For a two - tailed test with \(\alpha = 0.05\) and \(df=n - 1=6 - 1 = 5\). Looking up in the t - distribution table, the critical values are \(t=\pm 2.571\)

Step5: Make a decision

Since the calculated \(t = 4\) and \(|t|=4>2.571\) (the critical value), we reject the null hypothesis.

Answer:

The sample provides evidence of a significant treatment effect.