Dr. Yentz reads a report that claims the average college male drinks 9 alcoholic beverages a week. He wants to determine if males at the college he works at drink an amount that differs from this average. Based on a sample of 20 male students at his college he finds a mean of 7.6 with a standard deviation of 2.5. Conduct a non-directional one sample t-test using a significance level of .05 to determine if males at his college have an average alcohol consumption per week that differs from an average of 9 drinks per week. The null hypothesis is that males at his college have a mean that is equal to The alternative hypothesis is that males at his college have a mean that is not equal to Will there be one positive Tcrit, one negative Tcrit, or a positive and negative Tcrit? To answer type in positive, negative, or positive and negative. What is the absolute vale of Tcrit? What is the standard error (rounded to 3 decimal places)? What is the obtained t-score (Tobt) (round to 2 decimal places)? Would this result produce a p-value less than .05 (answer yes, no, or impossible to say) Should the null hypothesis be rejected (answer yes or no)? If Dr. Yentz uses the results to conclude that males at his college drink significantly less than 9 drinks a week is he accurately interpreting the results (answer yes or no)? Compute Cohen’s d for this analysis. Cohen’s d is (round to decimal 2 places)

To calculate Fest, we will first find Tcrit, the absolute T value, the standard error, Tobt, and then determine if the p-value is less than .05.

Degrees of freedom (df) = 20 - 1 = 19
Tcrit (two-tailed with alpha = .05) = ±2.093
Standard error = 2.5 / sqrt(20) = 0. the standard error is 0.560
Tobt = (7.6 - 9) / 0.560 = -3.214

Absolute value of Tobt = 3.214
The answer to the question is negative.

It is possible to calculate the p-value with the T value, degrees of freedom, and the above analysis. However, the result is not required to determine if the null hypothesis should be rejected. The p-value is less than .05, therefore the null hypothesis should be rejected.

Dr. Yentz would accurately interpret the results if he concludes that males at his college drink significantly less than 9 drinks per week based on these results.

To compute Cohen's d, we can use the formula:
Cohen's d = (mean of sample - mean of population) / standard deviation
Cohen's d = (7.6 - 9) / 2.5 = -0.76

Therefore, Cohen’s d for this analysis is -0.76.