A researcher sampled 25 participants and measured their alcohol consumption (in

Question

A researcher sampled 25 participants and measured their alcohol consumption (in ounces) in both a home and a bar setting. If the mean difference in satisfaction ratings for this sample was 11.4±2.5 (MD +/- SMD), then what is the decision at a .05 level of significance?

A)Alcohol consumption significantly differed, t(23) = 4.56.

B) Alcohol consumption significantly differed, t(24) = 4.56.

C) Alcohol consumption significantly differed, t(23) = 22.80.

D) Alcohol consumption significantly differed, t(24) = 22.80.

Explanation / Answer

Here the degrees of freedom of t-distribution = n - 1 = 25 - 1 = 24.

t(24) = 4.56.

Hence Option(B) is the correct choice. (Ans).

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