Aresearcher 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.

(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 be the true difference between testing averages for students using Method 1 and students using Method 2. We have the sample sizes and , the unbiased point estimate for is , i.e., 89.6 - 68.8 = 20.8

The standard error is given by , i.e.,

= 1.3810. Then, the endpoints for a 98% confidence interval for is given by

20.8-1.3810 and 20.8+1.3810, i.e.,

20.8-1.3810 and 20.8+1.3810 where is the the first quantile of the standard normal distribution, i.e., -2.3263, so, we have

20.8-(2.3263)(1.3810) and 20.8+(2.3263)(1.3810), i.e.,

17.5874 and 24.0126

answer: the resulting cross section is a rectangle. solution: a plane parallel to the base of a triangular prism will intersect a cross section that is the same shape as its bases. so the cross section is a triangle.

step-by-step explanation: