Bribe ‘em with Chocolate In a study published in the journal T...

Question

Bribe ‘em with Chocolate In a study published in the journal Teaching of Psychology, the article “Fudging the Numbers: Distributing Chocolate Influences Student Evaluations of an Undergraduate Course” states that distributing chocolate to students prior to teacher evaluations increases results. The authors randomly divided three sections of a course taught by the same instructor into two groups. Fifty of the students were given chocolate by an individual not associated with the course and 50 of the students were not given chocolate. The mean score from students who received chocolate was 4.2, while the mean score for the nonchocolate groups was 3.9. Suppose that the sample standard deviation of both the chocolate and nonchocolate groups was 0.8. Does chocolate appear to improve teacher evaluations? Use the α = 0.01 level of significance.

Accepted Answer

Hello. In this video, we are told that a researcher wants to know if providing free coffee before a morning exam improves student performance. In a randomized experiment, 60 students receive coffee and 60 do not. The mean exam score for the coffee group is 78 while the mean for no coffee group is 75. Both groups have a sample standard deviation of 6. We want to test whether the coffee improves exam scores at the 0.01 significance level. So let's go ahead and first write down our given information. We will define the group scores with coffee as new one, so these are going to be the average score. With coffee And we're going to define you too as the average score. Without coffee. Now the problem gives us the following information. We are told that 60 students receive coffee and 60 students do not receive coffee, so the sample size for the first group is going to be 60 and the sample size for the second group will be 60 as well. The mean exam score of the first group with coffee is going to be 78, and the mean exam score of the second group is going to be 75. And the standard deviation of the group that has coffee is 6, and the same can be said without the coffee. So both of the groups have the same sample standard deviation of 6. So, what we want to go ahead and do is we want to go ahead and test the claim of whether the coffee improves exam scores at the 0.01 significance level. So we are working with the significance level of 0.01. Now again, the claim states that if the students take coffee, then their exam scores will be greater than the students without coffee. So the claim here is that new 1 is going to be greater than U 2. Now in this case here we are testing the claim with two different populations with known standard deviations, and because we have known standard deviations we are going to set this claim as our alternate hypothesis. With that being said, our no hypothesis is going to be the complement of this claim, and the complement is that the group that takes coffee is going to have scores that are less than or equal to the group that doesn't take coffee. Furthermore, we are going to use a left-tailed, apologies, right-tailed test, because we are using the greater than symbol. So we have a right-tailed test for this problem. And on top of that, we are going to use a right-tailed tea test. So this is going to be our null and alternate hypothesis, and we know that we are going to be using a right-tailed T test. Now since we know that the populations and the standard deviations are equal to each other, what this means is that we are going to, we don't know anything about the sample variances, but because we are going to assume that we have unequal variances, we are going to approach this problem using Welch's T test. So, let's go ahead and define the T test statistic using Welch's T test. Now the T test statistic is going to be defined as the following. It's the average scores of the first group, minus the average scores of the 2nd group, divided by the square root of the 1st standard deviation squared, divided by the 1st sample size, plus the 2nd sample standard deviation squad, divided by the 2nd sample size. So we have all this information, let's go ahead and plug this in to find our T test statistic. So this is going to give us 78 minus 75 divided by the square root. Of 6 square divided by 60 plus 6 2 divided by 60 and with the help of a calculator this is going to give us an approximate t test statistic of 2.74. Now what we need to do is we need to find the critical value. Now, the critical value requires two things. We need the significance level, and we need the degree of freedom. But through the Welch's tea test, the degree of freedom is defined as the following. The degree of freedom is a complex fraction. In the numerator we have the quantity standard deviation of the first group squared divided by the first sample size, plus the standard deviation of the second group squared divided by the second sample size. This entire thing is squared, and it will be divided by the sum of two fractions. The first fraction is the standard deviation of the first group squared divided by the first sample size, quantity squared divided by 1 minus the first sample size, plus. The standard deviation of the 2nd group squared divided by the 2nd sample size squared. All divided by 1 minus the 2nd sample size. So this is a bit much. However, we do have all the information we need to find this degree of freedom. So By plugging in the known values, we will have the following degree of freedom. This is going to give us 6 squared. Divided by 60 plus 6 2 divided by 60, this entire race to the power of 2. Divided Divided by 62 divided by 60. Divided by 60 minus 1, which is 59, plus 6 2 divided by 60, all divided by 60 minus 1, which is 59, and with the help of a calculator we are going to get a degree of freedom of 118. So earlier, we were told to work with a significance level of 0.01, and we just calculated a degree of freedom of 118. So now we can go ahead and find our T critical value. Now, we are going to use a T critical value where the degree of freedom is 118 and the significance level is 0.01. And recall that we are using a right tail test so we don't have to change the significance level. So by using a t table, this is going to give us a T critical value of approximately 2.358. Now that we have the T critical value, we are going to compare this value to the T test statistic we calculated earlier. So let me go ahead and make some more space for that. OK, So earlier, we calculated that the T test statistic is going to be 2.74. And we have a T critical value of 2.358. Well, by taking a look at this value, we know that our t test statistic is going to be less than our t critical value. So what does this tell us about the null hypothesis? Well, this result tells us to reject. The null hypothesis. And since we are rejecting the null hypothesis, this tells us that there is sufficient evidence to conclude that providing coffee improves exam scores than not providing coffee, and the solution to this problem is going to be option A. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

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