Two researchers conducted a study in which two groups of...

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 26.8 with a standard deviation of 4.9, while the 200 students in group 2 had a mean score of 17.2 with a standard deviation of 3.7. Complete parts (a) and (b) below. (a) Determine the 90% confidence interval for the difference in scores, ?1 - ?2. Interpret the interval. The lower bound is. The upper bound is. (Round to three decimal places as needed.) Interpret the interval. Choose the correct answer below. A. The researchers are 90% confident that the difference of the means is in the interval. B. There is a 90% probability that the difference of the means is in the interval. C. The researchers are 90% confident that the difference between randomly selected individuals will be in the interval. D. There is a 90% probability that the difference between randomly selected individuals will be in the interval. (b) What does this say about priming? A. Since the 90% confidence interval does not contain zero, the results suggest that priming does have an effect on scores. B. Since the 90% confidence interval contains zero, the results suggest that priming does have an effect on scores. C. Since the 90% confidence interval does not contain zero, the results suggest that priming does not have an effect on scores. D. Since the 90% confidence interval contains zero, the results suggest that priming does not have an effect on scores.

Transcript

00:01 All right, in this test, we're given two different groups of students, each 200, or each with a size of 200 students in the group, which are given two different priming treatments before taking trivia or like some kind of examination, where the sample average of the first students who were told to think about what it means to be a professor is equal to 26 .8 with a standard deviation of 4 .9, and the average score of those who were thought to think about, i think, the question said soccer hooligans is 17 .2 with a standard deviation of 3 .7.

00:37 And we're asked to construct a 90 % confidence interval for the population difference in means.

00:42 And so to do this, we're going to need to calculate our t statistic for the difference in population means, which is going to be equal to our difference in sample means, minus our null hypothesis for the difference in population means, which our null hypothesis is going to be equal to zero.

01:02 That there is no difference.

01:05 And so i'm doing mu 1 -0 minus mu -2 -not, or i guess another way, a more clear way of putting, so it would be mu -1 minus mu -2, not for our null hypothesis, divided by the pooled standard deviation times the square root of 1 over n1 squared plus 1 over n2, or not squared, sorry, just 1 over n1, plus 1 over n2.

01:43 Hold on, someone just came to the door.

01:46 Sorry about that.

01:47 Okay.

01:48 So, and to calculate our t statistic in this particular case, this is going to be equal to, or actually, we don't even need to calculate the t statistic per se.

02:04 We just need to get the standard error of the sampling distribution, which is going to be this value right here, because we're constructing a confidence interval.

02:12 And so our standard error of the competence interval or of the test statistic is going to be sp times the square root of 1 over n1 plus 1 over n2.

02:24 And our confidence interval is going to be our observed difference in sample means plus or minus our critical value of t times the standard error of the sampling distribution, which is again equal to s or the pooled standard.

02:44 Standard deviation times the square root of 1 over n1 plus 1 over n2.

02:54 So we need to find our critical value of t.

02:57 And so to do this, we need to know how many degrees of freedom we have.

03:01 Our degrees of freedom is equal to n1 plus n2 minus 2, which in this case is going to be 400 minus 2, which is 398.

03:12 And so we can probably assume that this t distribution approximates a normal distribution.

03:23 So yeah, it's going to be somewhere between 100 and 1 ,000 degrees of freedom.

03:28 Let's use 100 because it's slightly more conservative.

03:34 And we are building a 90 % confidence interval, which means at a two -tailed, for two -tilled test, the significance level would be 0 .10.

03:44 Basically, this means that if this is our t distribution, at this value, 5 % of the mass, lies above this value of t and 95 here.

04:01 Cumulative probability lies below, 95 % of the mass lies below, which means our 90 % confidence interval is going to be between plus or minus this value of t critical.

04:14 And so our t critical value is going to be 1 .660.

04:25 So this is going to, our confidence interval is going to be 26 .8 minus, 17 .2 plus, what did i say it was, 1 .660.

04:42 And if you're worried about rounding or using a thousand degrees of freedom, et cetera, i'll do, i'll run through the case two where we use 1 .646 as are t critical, but for now we're going to use 1 .66.

04:58 So first we need to find our pooled standard deviation.

05:01 This is going to be equal to 4 .9 squared times 100 .5 .5 .5 .5 .5 .5.

05:09 299 plus 3 .7 squared times 199 divided by 398...

Question

Please give Ace some feedback