PSY 555 Homework 11

Chapter 7: #6,8,9,10,13,14,16,25 (for each problem involving t-tests, please list what type of test is appropriate to be used).

7.6(a).   North Dakota Verbal SAT:

z=

The probability of a z as extreme as , so we would reject the hypothesis that the SAT has a mean of 500 and a standard deviation of 100.

7.6(b).   Using a one-tailed (), the null hypothesis would not have been rejected because the difference (although significant by a two-tailed test), was in the unexpected direction.

7.8.      Arizona math SAT

z=

We would reject H0.

7.9.      The answer to Exercise 7.8 differs substantially from Exercise 7.6 because the sample sizes are so very different.  I deliberately sought examples where the means were nearly the same, but with that large difference in sample size, so the resulting z values, and associated probabilities, are very different.

7.10.     For the North Dakota verbal SAT data in Exercise 7.6:

CI.95==525 =525

512.3

The 95% confidence interval would be 512.3 to 537.7.

7.13(a). Performance when not reading passage

t=

7.13(b).  This does not mean that the SAT is not a valid measure, but it does show that people who do well at guessing answers also do well on the SAT.  This is not very surprising.

7.14(a)  Testing the experimental hypothesis that children tend to give socially-approved responses:

I would compare the mean of this group to the mean of a population of children tested under normal conditions.

7.14(b). The null hypothesis would be that these children come from a population with a mean of 3.87 (the mean of children in general).  The research hypothesis would be that these children give socially-approved responses at a different rate from normal children because of the stress they are under.

7.16.     Beta-endorphin levels:

Gain scores:

10 7.5 5.5 6 9.5 -2.5 13 3 -.1 .2 20.3 4 8 25 7.2 35 -3.5 -1.9 .1

Mean=7.70     SD=9.945

t=

Reject H0 and conclude that beta-endorphin levels were higher just before surgery.

7.25(a).  Everitt's Family Therapy Group:

We want to test the null hypothesis that the mean weight was the same before and after treatment.

t=

7.25(b). t=4.185 on 16 df, which tells us that there was a significant gain in weight over the course of therapy.

1.        Within subjects t-tests are used when the same participants are present in both groups being compared (i.e., a pre and post group, different drug conditions, etc.). Between subjects t-tests are used when different participants comprise the groups being compared (i.e., gender comparisons, most t-tests where you are looking for differences on a given categorical variable). The types of within subjects t–tests include difference score t-tests, matched sample t-tests, related sample t-tests, and repeated measures tests.  Between subjects t-test refer to those that are conducted on two independent samples.

2.        We need to test for homogeneity of variance because it is an assumption of the t-statistic.  If heterogeneity of variance exists, then the values in the t-distribution table will not accurately reflect the probability of occurrence in your sample.  If heterogeneity of variance exists, we need to make a correction to the degrees of freedom to ensure that our conclusions are sound.