Statistics concepts and applications

¶ … scores of first born and second born children on the Perceptual Aberration test. Ha: There is a significant difference in mean scores of first born and second born children on the Perceptual Aberration test. The alternative hypothesis is one sided, since the null hypothesis is concerning non-directional data, in that we are not predicting directional information.

The data will be examined using an independent t test. This test is used since the groups in these circumstances are not related. If the samples were correlated, where each individual had two scores under two treatment conditions, the dependant t test would have been used.

The test statistic in this experiment represents the difference between the mean scores of the children tested divided by the standard error of the difference. This result would represent whether there was a significant difference between the means. If so, we would reject the null hypothesis. By using the t test, we are able to determine the ratio of the mean difference in test scores when compared to the error of differences in the means. A large mean difference does not guarantee a large t, hence the use of the standard error of difference.

1d. Following a .05 level of significance, and after calculating the df (92), the critical value needed to reject the null hypothesis is 2.367. Our calculations show that t=.529982. Thus, we would accept the null hypothesis because our calculated value (t = .529982) is less extreme than the critical values (2.367 or -2.367).

1e. The observed mean difference in test scores between first (M=17.2563) and second (M=16.14815) born children was not significantly different, t (92) = .529982, p>.05.

t =

17.2963-16.14815

4331.62 +1497.407

1

1

92

27

27

t = 1.14815 / SQRT (5829.037/92) * ((1/27) + (1/27))

t =

1.14815 / SQRT (63.3591 *.074074) = 1.14815 / 2.166395 = .529982

2a. H0: There is no difference in the population means of the different scores on recall of words testing following memory techniques. Ha: There is a difference in the population means of the different scores on recall of words testing following memory techniques.

2b. An independent two samples t test is not appropriate because our samples include more than two samples. We are comparing the results of three groups of subjects.

2c. The between groups ANOVA test statistic will measure if there is a difference between the mean scores of subjects on recall of words testing following the use of memory techniques. In short, it measures the variance among the means of the three populations to determine whether or not there is a significant difference.

2d. Following a .05 level of significance, and after calculating the df (2), the critical value needed to reject the null hypothesis is 3.05. Our calculations show that F=48.2131. Thus, we would reject the null hypothesis because our calculated value (F=48.2131) is larger than the critical F. value of 3.05. Thus, there is a difference in the population means of the different scores on recall of words testing following memory techniques.

2e. The researcher hypothesized that there was a difference in the population means of the different scores on recall of words testing following memory techniques. This hypothesis was supported by finding that the calculated F. value (48.2131) was higher than the critical F. value of 3.05, p

Summary Table Showing Values

Source

df

SS

MS

F

p

IV

2

48.2131

Error

9.319529

Total

Group 1: Word Association

Group 2: Stress technique

Group 3: No memory technique

2g. The purpose of adding the third group, or that of the control group, was to ensure internal validity. This concept refers to the idea that, by using a control group, a researcher can ensure that the sampling method, testing method, and randomization method did not create a threat to the overall results of the test.