Question

A researcher conducts a study on the effects of amount of sleep on creativity. The creativity scores for four levels of sleep (2 hours, 4 hours, 6 hours, and 8 hours) are presented below:

            2 Hours of Sleep         4 Hours of Sleep         6 Hours of Sleep         8 Hours of Sleep

                        3                                  4                                  10                                10

                        5                                  7                                  11                                13

                        6                                  8                                  13                                10

                        4                                  3                                  9                                9

                        2                                  2                                  10                                10

Pretend that you have only the 4-hour group of data and the 6-hour group of data. Also pretend that this data set does not meet all the conditions for parametric testing. Without using a parametric procedure, compare the two samples (4-hour and 6-hour samples) and test for a significant (p < 0.05) difference.

(a) What is the appropriate null hypothesis for this situation? What is the alternative hypothesis?

(b)Identify and perform the most appropriate procedure to test the null hypothesis, present your

      results, and draw your conclusion using a complete sentence.

Answer 1

Solution:-

a)

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

Null hypothesis: u₁ = u₂
Alternative hypothesis: u₁ u₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 1.342
DF = 8
t = [ (x₁ - x₂) - d ] / SE

t = - 4.32

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than -4.32; that is, less than -4.32 or greater than 4.32.

b)

Thus, the P-value = 0.003.

Interpret results. Since the P-value (0.003) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is difference between the the two samples (4-hour and 6-hour samples).