A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 175 students using Method 1 produces a testing average of 89.6. A sample of 177 students using Method 2 produces a testing average of 68.8. Assume the standard deviation is known to be 17.56 for Method 1 and 5.07 for Method 2. Determine the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Answer:(17.5874, 24.0126) is the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.Step-by-step explanation:Let [tex]\mu_{1}-\mu_{2}[/tex] be the true difference between testing averages for students using Method 1 and students using Method 2. We have the sample sizes [tex]n_{1} = 175[/tex] and [tex]n_{2} = 177[/tex], the unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 89.6 - 68.8 = 20.8The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}[/tex], i.e.,[tex]\sqrt{\frac{(17.56)^{2}}{175}+\frac{(5.07)^{2}}{177}}[/tex] = 1.3810. Then, the endpoints for a 98% confidence interval for [tex]\mu_{1}-\mu_{2}[/tex] is given by20.8-[tex]z_{0.02/2}[/tex]1.3810 and 20.8+[tex]z_{0.02/2}[/tex]1.3810, i.e.,20.8-[tex]z_{0.01}[/tex]1.3810 and 20.8+[tex]z_{0.01}[/tex]1.3810 where [tex]z_{0.01}[/tex] is the the first quantile of the standard normal distribution, i.e., -2.3263, so, we have20.8-(2.3263)(1.3810) and 20.8+(2.3263)(1.3810), i.e.,17.5874 and 24.0126