In Exercises 5–20, assume that the two samples are independent...

Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)

Readability of Font On a Computer Screen The statistics shown below were obtained from a standard test of readability of fonts on a computer screen (based on data from “Reading on the Computer Screen: Does Font Type Have Effects on Web Text Readability?” by Ali et al., International Education Studies, Vol. 6, No. 3). Reading speed and accuracy were combined into a readability performance score (x), where a higher score represents better font readability.

a. Use a 0.05 significance level to test the claim that there is no significant difference in readability between Roman and Arial fonts.

Accepted Answer

Welcome back, everyone. In this problem, a researcher wants to determine if there is a significant difference in math test performance between students who use digital textbooks and those who use printed textbooks. The following data were collected. Here we have the groups to sampleize their means and their standard deviations. Assume that the two samples are independent simple random samples selected from normally distributed populations and do not assume that the population standard deviations are equal. Use degrees of freedom equal to the smaller of N1 minus 1 and N2 minus 1. At the 0.05 significance level, test the claim that there is no significant difference in mean test performance between the two groups. Now for our hypothesis test here, let's state our hypothesis to help us test our claim, OK? So now, the claim is that there is no significant difference in mean test performance, OK? So that means the null hypothesis. I So another hypothesis, it's not, is that the means do not differ. In other words, the mean of the first equals the mean of the second. There is no difference. Which means then that our alternative hypothesis H1 says that there is a difference, that mu1 is not equal to mu 2, OK. So, no, because we're testing whether two values are equal to each other or not, then we are going to use a two-tailed test. And in this case, we'll use a two-tailed. Tea test, OK. And we'll use a 2 sample T test because of the unequal variances. Now, what are we really trying to figure out here? Well, OK, since we are using our T test. Right, uh, what we're really saying is that if we were to represent our distribution, OK. Then we want to find the likelihood that our test statistic is going to fall in between the absolute value for the critical value that we find, OK? So whatever critical value that we find in this case, this will be the T value for half of alpha and the degrees of freedom, OK. Then we're saying that we want to test. OK, if our test statistic is within this region. And now we know that if it is within this region, OK. If it's within this region, then we can tell. That we can, we would fail to reject the null hypothesis, OK. So we failed to reject the non-hypothesis, but then if it is outside of the region, then we would reject the non hypothesis, OK? So now, let's go ahead and first try to find the test statistic. Now recall that for the two sample tests with unequal variances, the test statistic. It's gonna be equal to the difference between our. Difference between our means as X1 minus X2, OK. Minus 1 minus m2, which is often assumed to be 0, divided by the square root of our standard deviation squared divided by n1, which is the sample size for a first value, plus the standard deviation of the second value square divided by N2. Knowing our problem, we're given all of this information because we know that for the digital group, if we label it as the first group, and one. Will be equal to 30. X bar 1 is equal to 78.4 the mean, and the standard deviation S1 equals 10.2. On the other hand, for our printed group. If we label that as the second, then N2 is 28. OK. Uh X bar 2 is 81.1 and S2 is equal to 12.5. So now by substituting these values. Then we should get our test statistic to be equal to 78.4 minus 81.1, OK, minus 0. Like we said, we often assume that to be zero, divided by the square root, OK, of 10.2 squared divided by 30. Thus 12.5 square divided by 28. Now, when we go ahead and solve, we should get our test statistic to be approximately equal to 0.898. Now that we have our test statistic, we need to find our critical value, OK? Now, let me put that in blue here. Now we know that like the problem told us, the degrees of freedom will be the least of value depending on the sample size. So it's gonna be the minimum value of N21 minus 1, 30 minus 1. And the minimum value of N2 minus 1, 28 minus 1. That's a minimum value of 29 and 27, which would be equal to 27. And now, for a two-tailed test, OK, for a two-tailed test. At Alpha equals 0.05 and 227 degrees of freedom. OK. Then our critical value, OK, 0.025 and 27, is gonna be approximately equal to 2.052. So what does that mean? Well, if we go back to our, uh, little diagram here, that means that we would fail to reject the hypothesis if it's between -2.052 and 2.052 on our uh T test on our graph. And in this case, since we found it to be -0.898, which would probably fall somewhere right here, OK. Then we can say that let me put this in a box to the right. We can say that since the absolute value of T is equal to 0.898, which is less than 2.052, OK, then we fail to reject the null hypothesis. OK. So thus, by our conclusion, at the 0.05 significance level, there is not enough evidence to conclude that there is a significant difference in mean math test performance between students using digital and printed textbooks. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Independent Samples

Hypothesis Testing

T-test for Independent Samples

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