Before the two months of journaling, the mean score on the impact of event scale

Question

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 14.5. After the two months of journaling, the mean score was 14.8. The mean of the differences between each person's pre- and post- scores was 0.3, with a standard deviation of the differences equal to 1.4. The graduate student has no presupposed assumptions about whether journaling can affect grief and healing, so she formulates the null and alternative hypotheses as: H_0: mu_0 = 0 H_1: mu_0 notequalto 0 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 = 49 = 1 = 48 degrees of freedom. This is a ____ test. Use the Distributions tool to find the critical score(s) for the level of significance alpha = .05. The critical score(s) (the value(s) for t that separate(s) the tail(s) from the main body of the distribution, forming the critical region) is/are ____. To calculate the test statistic, you first need to calculate the estimated standard error under the assumption that the null hypothesis is true. The estimated standard error is ____. The test statistic is t = ____. use the tool to evaluate the null hypothesis. The t statistic ____ in the critical region for a two-tailed hypothesis test. Therefore, the null hypothesis is _____. The graduate student _____ conclude that journaling affects grief. The graduate student repeats her study with another random sample of the same size. This time, instead of the treatment being journaling about their emotions, the treatment is attending a support group for widows and widowers. Suppose the results are very similar. After attending a support group for widows and widowers, the mean score was still 0.3 higher, but this time the standard deviation of the difference was 1.1 (vs. the original standard deviation of 1.4). This means _____ has the more consistent treatment effect. This difference In the standard deviation also means that a 95% confidence Interval of the mean difference would be for the original study, when the treatment was journaling about their emotions, than the 95% confidence Interval of the mean difference for the second study, when the treatment was attending a support group for widows and widowers. Finally, this difference in the standard deviation means that when the graduate student conducts a hypothesis test testing whether the mean difference is zero for the second study, she will be _____ likely to reject the null hypothesis than she was for the hypothesis test you completed previously for the original study.

Explanation / Answer

Solution:-

xd = 0.3

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

Null hypothesis: d = 0

Alternative hypothesis: 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.

a) 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).

b) tcritical = + 2.011

s = sqrt [ ((di - d)2 / (n - 1) ]

s = 1.4

c) SE = s / sqrt(n)

S.E = 0.2

DF = n - 1 = 49 -1

D.F = 48

t = [ (x1 - x2) - D ] / SE

d) t = 1.5

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.

The test statistics does not lie in the critical region for a two tailed hypothesis test. Therefore the null hypothesis is true. The graduate students does not conlude that journaling affects griefs.

This means that this new test has more consistent effect.

The difference in the standard deviation also means that a 95% confidence interval of the mean difference would be not significant for the original study.

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