Question

A researcher investigates whether or not a new cold medication disrupts mental alertness. It is known that scores on a standardized test containing a variety of problem-solving tasks are normally distributed with µ = 64 and σ = 8. A random sample of n = 16 subjects are given the drug and then tested. For this sample, the mean is M = 58. a. Are the data sufficient to conclude that the medication affects performance? Test with α = .01, two tails. b. Compute Cohen's d to measure the size of the treatment effect

Answer 1

Given: = 64, = 58, = 8, n = 16, = 0.01

The Hypothesis:

H0: = 64

Ha: 64

This is a 2 tailed test

The Test Statistic: Since n < 30, we use the students t test.

The test statistic is given by the equation:

t = \frac{\bar{x}-\mu }{\frac{\sigma}{\sqrt{n}}} = \frac{58-64}{\frac{8}{\sqrt{16}}} = -3

t observed = -3

The p Value: The p value(2 tailed) for t = -3, for degrees of freedom (df) = n-1 = 15, is;

p value = 0.0091

The Critical Value:   The critical value (2 Tail) at = 0.01, for df = 15, tcritical = +2.947 and -2.947

The Decision Rule:   If tobserved is > tcritical or if tobserved is < -tcritical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision:   Since tobserved (-3) is < tcritical (-2.947), We Reject H0.

Also since P value (0.0091) is < (0.01) , We Reject H0.

The Conclusion: There is sufficient 99% significance level to conclude that the medication affects performance.

(b) Cohens d = (Sample Mean - Population Mean)/Standard deviation = 58-64/8 = -6/8 = -0.75