Question

5. A two-tailed hypothesis test for a repeated-measures design A graduate student is interested in whether jounaling can affect grief and healing. For her study, she selects a random sample of 64 adults who have been widowed within the past 10 to 14 months. The subjects complete a battery of psychological questionnaires before and after spending two months journaling about their emotions. Before the two months of journaling, the mean score on the impact of event scale-avoidance subscale, which measures how much individuals consciously avoid thoughts and feelings associated with the loss of their spouses, was 15.4. After the two months of journaling, the mean score was 15.0. The mean of the differences between each person's pre- and post-scores was 0.4, with a standard deviation of the differences equal to 1.9 The graduate student has no presupposed assumptions about whether journaling can affect grief and healing, so she formulates the null and alternative hypotheses as: She uses a repeated-measures t test. Because the sample size is large, if the null hypothesis is true as an equality, the test statistic follows a t-distribution with n-1-64-1-63 degrees of freedom. Degrees of Freedom-55 esc F2 F3 F5 8 4

Answer 1

SOLUTION

Let X = Score before spending 2 months journeling.

      Y = Score after spending 2 months journeling.

And D = X - Y.

Then, D ~ N(µ, σ²) where σ² is unknown.

Hypotheses:

Null: H₀: µ = µ₀ = 0 Vs Alternative: H_A: µ ≠ 0

Test Statistic:

t = (√n) (Dbar)/s where

Dbar and s are respectively, sample average and sample standard deviation based on n observations on X and Y.

Calculations

t = (√64) (0.4)/1.9

= 1.684

Distribution, Critical Value and p-value:

Under H₀, t ~ t_{n - 1}. Hence, for level of significance α%, Critical Value = upper (α/2)% point of

t_{n - 1} and p-value = P(t_{n - 1} > | tcal |).

Taking α = 0.05, using Excel Functions, the above are found to be as follows:

tcrit = 1.998 and p-value = 0.0971

Decision:

Since | tcal | < tcrit, H₀ is accepted. Or equivalently, since p-value > α, H₀ is accepted.

Conclusion:

There is sufficient evidence to suggest that the jounaling does not impact the score. ANSWER

DONE