Question

1. Is energy drink X better than a placebo when it comes to concentration? To find out, two randomly divided sets of participants (15 in each group) are given one of the two: either drink X or the placebo. Each participant is then asked to take a certain test; the higher the score, the better their concentration. After the scores are recorded, it’s found that those on drink X scored an average of 76.5 with a standard deviation of 4.9, and those on placebo scored an average of 69.5 with a standard deviation of 5.2.

   1. What kind of test is this? What are the hypotheses?

   2. Check the conditions.

   3. Run the test and find the p-value. Then give your conclusion in context.

Answer 1

Since we are comparing the mean scores of two samples and we do not know the population standard deviations, we will use two-sample t test.

Let and be the average score for drink X and placebo group.

Null hypothesis H0: Mean scores recorded for drink X and placebo are equal. That is =

Alternative hypothesis Ha: Mean scores recorded for drink X is greater than placebo. That is >

We need to check for homogeneity of variances.

Test statistic, F = s2² / s1²   = 5.2² / 4.9²  = 1.12

Numerator df = n2 - 1 = 15 - 1 = 14

Denominator df = n1 - 1 = 15 - 1 = 14

Critical value of F at 0.05 significance level and df = 14, 14 is 2.48

Since the observed F (1.12) is less than the critical value, we conclude that there is no significant evidence that the population variance (or standard deviations) of the two groups are not equal.

So, we assume that the population standard deviations of both groups to be equal and we can apply pooled t-test.

Pooled variance Sp² = [(n1 - 1) s1²   + (n2 - 1)s2² ] / (n1 + n2 - 2)

= [(15 - 1) 4.9² + (15 - 1) ] 5.2² / (15 + 15 -2)

= 25.525

Standard error of mean difference = sqrt( Sp² * ((1/n1) + (1/n2)))

= sqrt( 25.525* ((1/15) + (1/15)))

= 1.844813

Test statistic, t = (x1 - x2) / Std error

= (76.5 - 69.5) / 1.844813

= 3.79

Degree of freedom = n1 + n2 - 2 = 15 + 15 - 2 = 28

P-value = P(t > 3.79, df = 28) = 0.0004

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that mean scores recorded for drink X is greater than placebo.