Question

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 116 students using Method 1 produces a testing average of 50.3. A sample of 151 students using Method 2 produces a testing average of 68.6. Assume that the population standard deviation for Method 1 is 15.62, while the population standard deviation for Method 2 is 17.21. Determine the 99% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval.?

Answer 1

n1 = 116, n2=151,

= 50.3,  = 68.6

= 15.62,  =17.21

C= 99%

Formula for confidence interval is

Where Zc is the Z critical value for C= 99%

Critical value = 2.576 2.58

Critical value = 2.58

-18.3 5.201626739

-23.49 < < -13.11

thus confidence interval is ( -23.49 , -13.11)