In a study conducted to determine the role that sleep disorder...

Question

In a study conducted to determine the role that sleep disorders play in academic performance, researcher Jane Gaultney conducted a survey of 1845 college students to determine if they had a sleep disorder (such as narcolepsy, insomnia, or restless leg syndrome). Of the 503 students with a sleep disorder, the mean grade point average was 2.65 with a standard deviation of 0.87. Of the 1342 students without a sleep disorder, the mean grade point average was 2.82 with a standard deviation of 0.83. Source: SLEEP 2010: Associated Professional Sleep Societies 24th Annual Meeting.
b. Is there evidence to suggest sleep disorders adversely affect one’s GPA at the α=0.05 level of significance?

b. Is there evidence to suggest sleep disorders adversely affect one’s GPA at the α=0.05 level of significance?

Accepted Answer

Hello. In this video, a university wants to know if students who work part-time jobs have lower average GPAs than those who do not work. Out of 620 students who work part-time, the mean GPA is 2.71, and the standard deviation of 0.81. Among 980 students who do not work, the mean GPA is 2.87 with a standard deviation of 0.79. At a 0.05 significance level, is there evidence that working part time is associated with lower GPA. So, let's go ahead and first write down the given information. Let's define new one as the GPA. Of part-time workers. And we're going to define you too as the GPA. Of non-working students. Now the problem tells us that for the part-time workers, we have a sample size of 620. Their average GPA is going to be 2.71, and they have a sample standard deviation of 0.81. For the group that doesn't work, we have a sample size of 980. We have an average GPA. Of 2.87 And we have a sample standard deviation. Of 0.79 Now, we want to test whether the claim of working part-time is associated with lower GPAs. So the claim here is that the GPA of working students is going to be less than the GPA of non-working students. Now here, we are testing the hypothesis or we're doing a hypothesis test between two samples with known sample standard deviations, and because we know the sample standard deviations, this claim is going to act as our alternate hypothesis. And that means that the null hypothesis is going to be the complement of this claim, and the compliment is that the average GPA of part-time students is greater than or equal to the average GPA of non-working students. So Here we know. So here we are going to do a two sample t test for comparing means of independent groups, but we don't know anything about the variances, so we are going to assume unequal variances so we are going to use a two-tailed. Tea test. But more specifically, because we are using the less than symbol in our alternate hypothesis, we are specifically going to focus on the left side test. So this is going to be a left sided t test. And again, because we are working with unequal variances, or at least assuming we have unequal variances, we are going to approach this using Welch's T test. So, let's go ahead and start this problem by defining the T test statistic for Welch's T test. Now the T test statistic for Welch's T test is defined as the following. It is the 1st mean minus the 2nd mean divided by the square root of the 1st standard deviation squared divided by the 1st sample size, plus the 2nd sample standard deviation squared divided by the 2nd sample size. Now here the problem gives us all the information we need to calculate this test statistic. So by plugging in the known information, we will have 2.71 minus 2.87, and this is going to be divided by the square root of 0.81 squared. Divided by 620 plus 0.79 square divided by 980 and with the help of a calculator this is going to give us an approximate t test statistic of -3.89. So let's go ahead and set this to the side for now. The next thing we need to do is we need to calculate the critical value. Now, in order to calculate the critical value, there are two things that we need. We need the degree of freedom and the significance level, which the problem does give us. However, the degree of freedom through Welch's T test is defined as the following. It is a complex fraction. In the numerator we have the standard deviation of the 1st group squared divided by the 1st group sample size plus the standard deviation of the 2nd group squared divided by the 2nd sample size all squared. And this is going to be divided by the sum of two fractions. The first fraction's numerator is going to be the standard deviation of the first group squared, divided by the first sample size, quantity squared, all divided by the first sample size minus 1. Then we are going to add the standard deviation of the 2nd group squared divided by the 2nd sample size quantity squared, all divided by the 2nd sample size minus 1. So this is a bit much, but we do have all the information we need in order to calculate this degree of freedom. So by plugging in our known information, our degree of freedom is going to be defined as the following. It is 0.81 quantity squared divided by 620 plus 0.79 quantity squared divided by 980. This entire thing will be squared. All divided by 0.81 quantity squared divided by 620 this quantity squared. Divided by the first sample size minus 1, which is 619. Then we will add the second standard deviation squared, which is 0.79 squared, divided by the second sample size, which is 980. This quantity is squared, all divided by the 2nd sample size minus 1, which is 979. Now, with the help of a calculator, this is going to give us an approximate degree of freedom of 1,292. Now that we have the degree of freedom and recall that the problem does tell us to work with a significance level of 0.05, and we are using a left-sided test, let's go ahead and calculate the critical value. So the T critical value. Again, is using a left-sided test. We are working with a significance level of 0.05, and we have a degree of freedom of 1,292. By using a t critical table, this is going to give us a t critical value of approximately 1,645. Now, what we're going to do is we're going to take this critical value, and we're going to compare it to the T test statistic that we calculated earlier. Now, if we take a look, the T test statistic is less than the T critical value, so the T test statistic is less than the T critical value. So what does this tell us about the clean? Well, this tells us to reject the null hypothesis. So we are rejecting the null hypothesis, but what does this tell us about the original claim? Well, this tells us that there is sufficient evidence to conclude that working part time is associated with lower GPAs. And with that being said, 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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