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.

b. Construct the confidence interval suitable for testing the claim in part (a).

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 our group sample size mean and 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. Construct a 95% confidence interval for the difference in mean math test performance where the difference is defined as the mean for the digital textbook group minus the mean for the printed textbook group. Now, essentially, what are we trying to find here? Well, we want a 95% confidence interval for the difference in mean math test performance. What do we know about confidence intervals? Well, recall that the confidence interval for the difference in means for independent samples with unequal variances is X 1 minus X2. OK. The first mean minus the second mean plus R minus T multiplied by the standard error, OK? Where the standard error. Is equal to the square root. Of the first standard deviation squared divided by the first sample size plus the 2 standard deviation squared. Divided by the 2nd sample size. Now, in this case, for our data that we're given, OK, we know that for our digital group. The sample size N1 is 30, if we consider it the first group. The first mean X1 is 78.4, and its standard deviation is 10.2. On the other hand, for our printed group, if we consider it the second, then its sample size is 28, mean 81.1, and standard deviation, 12.5. OK? So no, we need to plug these values into our formula to find the difference in sample means and thus be able to solve, OK? So now, in this case. OK, in this case. We know right that here. Let's let me put this in red. The difference between sample means, OK. Is going to be equal to. X 1 minus X2, 78.4 minus 81.1, which is -2.7. Next, we can try to find uh our standard error of the difference. OK, and by the formula that means it will be equal to the square root of 10.2 square divided by 30. Plus 12.5 square divided by 28. And in New York that thought, we should get the standard error to be approximately 3.008. What else do we need? Well, no, we need the criticalt value for a 95% confidence interval. And now, in this case we know that it's a two-tailed test. OK, so first. Yeah, well, that's right, and here. We also know that 4. OK, uh, to tell the test. For a two-tailed tea test, actually. OK. With Alpha, since it's a 95% confidence interval, then alpha is 0.055%. And remember the degrees of freedom was the smaller of our two samples, 1 minus 1 and N2 minus 1. So the degrees of freedom equals the minimum value. Of 30 minus 1 and 28 minus 1, and we know that's 29 and 27, so our degrees of freedom is 70 is 27. Therefore, the critical T value, OK, for half of our uh value of alpha, 0.025 and our degrees of freedom 27 from our table of T values is going to be equal to 2.052. So let's do a quick recap. No, we have the difference between our sample means, we have our standard error of the difference, and we have the T, the value for our critical T value. OK? So, no, that means our or Our confidence interval then. It's going to be equal to -2.7 plus R minus T 2.052 multiplied by our standard error 3.008. In that case, OK, let me just split them up here. This is gonna be -2.7 plus or minus 6.2, which that means then the lower limit. Is going to be -2.7 minus 6.2, which is -8.9. While the upper limit. Is going to be -2.7 plus 6.2, which is 3.5. Thus, the 95% confidence interval for the difference in mean math test performance is from -8.9 to 3.5. Thanks a lot for watching everyone. I hope this video helps.

Key Concepts

Confidence Interval

Independent Samples

Standard Deviation

Watch next