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 (M_D ± S_MD),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.

Question

Quizplus · Accepted Answer

To determine the correct answer, we need to calculate the t-statistic using the given formula:
t = (mean difference - hypothesized difference) / (standard error of the difference)

The hypothesized difference is 0 (no difference between alcohol consumption at home and in a bar setting).

The standard error of the difference is calculated as:

SE = sqrt[ (s1^2/n1) + (s2^2/n2) ]

Where s1 is the standard deviation of the home consumption data, n1 is the sample size for home consumption, s2 is the standard deviation of the bar consumption data, and n2 is the sample size for bar consumption.

We don't have the actual data for calculating these values, but we are given that the mean difference is 11.4 ounces with a standard deviation of 2.5 ounces, and the sample size is 25. Therefore,

SE = sqrt[ (2.5^2/25) + (2.5^2/25) ] = 1.58

Plugging this and the other values into the t formula, we get:

t = (11.4 - 0) / 1.58 = 7.22

This t-value has degrees of freedom (df) equal to n - 1, or 24 in this case. We can look up the critical t-value at a 0.05 level of significance and 24 df in a t-table or calculator, which is approximately 2.064.

Since our calculated t-value (7.22) is greater than the critical t-value (2.064), we can reject the null hypothesis and conclude that there is a significant difference between alcohol consumption at home and in a bar setting. Therefore, the correct answer is B.

A Researcher Sampled 25 Participants and Measured Their Alcohol Consumption