SESN+9+ASSIGN-3

Fall 2014-SESSION 9 LECTURE and ASSIGNMENTS-3

Assignment 3: Revisiting our data from last week. Looking for an attribute variable. Discussion: Let’s look at the results from our study from last week. Notice in the data here that there are __19 in each group__, __Remember that these students are the same students under two different conditions__. Because these scores are from __one group of students__ we chose a paired t test The first piece of information tells whether the results are statistically significant—__p.__

What is probability that this could happen by chance alone? We see that the probability is greater than >05 so the results are not significant. In fact if we look at the number .83 (rounded off) we see that 83 times out of 100 these results would happen by chance alone.

That means we would only be about 17% sure of our results. Our treatment did not result in big enough differences between the two groups to show that the intervention is better than the more traditional approach even though the traditional lecture mean was a bit higher.

Sometimes we are trying something new, and the achievement differences are not significantly different but attitude is. We are happy that our new way of teaching does not negatively affect scores and see that the kids’ attitudes are significantly better when we use this new method. In this case statistical significance in achievement differences was not needed as long as the kids in the treatment group did not perform significantly lower.

The next paragraph on the quickcalcs result is not important to us for this discussion.

The third paragraph describes the t value and degrees of freedom. T, as explained in your chapters, is a numerical ratio comparing the scores in each group including the number of folks in each group and how spread out they are. Notice the letters df. That means degrees of freedom. That refers to the amount of folks in the study- some number depending on number of groups. Notice in this case it is 18 n-1 since there were 19 students and 1 group (same class looked at twice). The printout will supply you with the degrees of freedom for your study. If your analysis is a t test for paired samples the df would be n-1. Luckily the computer program figures all this out for you.

In your readings, there is talk about the critical value and whether t exceeded it. to determine the probability. This program skips over the critical value and gives you the probability. It makes it easier as you do not need to look it up on a chart.

Look all the way down at the summary table. This is very important. You see that the traditional method group had a higher mean score but not different enough to be significant. The standard deviation (sd) is smaller in the film group, which infers that there was a smaller range of scores and that the scores were more homogenous—good news!

Assignment 3 cont. Activity to be posted As mentioned in the Session’s preview, you may have noticed that gender may be working as an attribute variable. It looks like film might be better for boys than girls. If this is so, the gains for the boys are subdued by the losses of the girls. If we use gender as an attribute variable we can test this hypotheses. Conduct another two t tests—separate the tests by gender. Look at which condition the　girls did better and which condition the boys did better. Look at the differences in performance with respect to gender. Would you agree that gender is a factor? Explain. Post on discussion board with your research group.

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=ーー　答え　ーー= We checked their mean then we found that male students responded positively to the film treatment but female responded just the opposite as we mentioned below. This shows that this treatments have different effect to boys and girls. gender. However the differences are statistically non significant as the 95% confidence levels are met.

The means for each gender.

tradition mean film mean male -- 9 students 84.55 88.11 female -- 10 students 89.8 86.0

Girls average declined from 89.8 to 86 which is -3.8%. Boys average increased from 84.55 to 88.11 which is +3.56% Both cases it seems that the statistically significance in consideration of a 95% confidence levels are met.

=ーー　答え　終了ーー=

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