Question

A research study examined the effects of meditation on quality of sleep. Fifty individuals were asked how many nights, on average, they lost sleep at the beginning of the study. Then all fifty individuals were placed into a meditation intervention every day for three months. At the end of the three months, the same fifty individuals were asked how many sleepless nights they had on average. The mean difference between the before and after intervention groups was 5.6 nights, with a standard deviation of 5. Calculate the z, t, or chi-square test and choose the value in the answer below.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d< 0

Alternative hypothesis: u_d > 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

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 = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 5

SE = s / sqrt(n)

S.E = 0.70711

DF = n - 1 = 50 -1

D.F = 49

t = [ (x₁ - x₂) - D ] / SE

t = 7.92

where d_i 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 two-tailed test, the P-value is the probability that a t statistic having 49 degrees of freedom greater than 7.92.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.