Question

1. a researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(20) = 2.09. a. how many individuals participated in the entire study? b. using a two-tailed test with α = .05, is there a significant difference between the two treatments? c. using a two-tailed test with α = .01, is there a significant difference between the two treatments? d. compute r² to measure the percentage of variance accounted for by the treatment effect.

Explanation:

Part a

Step1: Recall df for independent - measures t - test

For an independent - measures t - test, the degrees of freedom formula is \(df = n_1 + n_2-2\), where \(n_1\) and \(n_2\) are the sample sizes of the two groups. We are given that \(df = 20\) (from \(t(20)=2.09\)).

Step2: Solve for total number of participants

Let \(N=n_1 + n_2\). From \(df=n_1 + n_2 - 2\), we can solve for \(N\). We know that \(df = 20\), so \(20=N - 2\). Adding 2 to both sides of the equation, we get \(N=20 + 2=22\).

Step1: Find critical t - value

For a two - tailed test with \(\alpha = 0.05\) and \(df = 20\), we look up the critical t - value in the t - distribution table. The critical t - value for \(df = 20\) and \(\alpha=0.05\) (two - tailed) is \(t_{crit}=\pm 2.086\) (approximate value from t - table).

Step2: Compare obtained t and critical t

The obtained t - statistic is \(t = 2.09\). Since \(|t|=2.09>t_{crit}=2.086\) (the absolute value of the obtained t is greater than the critical t - value), we reject the null hypothesis.

Step1: Find critical t - value

For a two - tailed test with \(\alpha = 0.01\) and \(df = 20\), we look up the critical t - value in the t - distribution table. The critical t - value for \(df = 20\) and \(\alpha = 0.01\) (two - tailed) is \(t_{crit}=\pm 2.845\) (approximate value from t - table).

Step2: Compare obtained t and critical t

The obtained t - statistic is \(t = 2.09\). Since \(|t|=2.09

Answer:

22

Part b