In: Statistics and Probability
In a experiment on relaxation techniques, subject's brain signals were measured before and after the relaxation exercises with the following results:
Person | 1 | 2 | 3 | 4 | 5 |
Before | 38.6 | 41.2 | 37.9 | 38.6 | 42.6 |
After | 37.0 | 37.0 | 34.3 | 36.9 | 43.6 |
Is there sufficient evidence to suggest that the relaxation
exercise slowed the brain waves? Assume the population is normally
distributed.
Select the test statistic, p-value, Decision to Reject
(RH0) or Failure to Reject (FRH0)].
a) [t = 2.214, p-value = 0.023, RH0]
b) [t = 1.127, p-value = 0.046, FRH0]
c) [t = 2.214, p-value = 0.091, FRH0]
d) [t = 1.127, p-value = 0.091, RH0]
e) [t = 2.214, p-value = 0.046, RH0]
Solution;- e) [t = 2.214, p-value = 0.046, RH0]
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: ud = 0
Alternative hypothesis: ud > 0
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
s = 2.04
SE = s / sqrt(n)
S.E = 0.9124
DF = n - 1 = 5 -1
D.F = 4
t = [ (x1 - x2) - D ] / SE
t = 2.214
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a one-tailed test, the P-value is the probability that a t statistic having 4 degrees of freedom is greater than 2.214.
Thus, the P-value = 0.046
Interpret results. Since the P-value (0.046) is less than the significance level (0.05), we have to reject the null hypothesis.
Reject H0. There is sufficient evidence to suggest that the relaxation exercise slowed the brain waves.