Question

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 214 students using Method 1 produces a testing average of 53.9. A sample of 193 students using Method 2 produces a testing average of 88.9. Assume that the population standard deviation for Method 1 is 17.61, while the population standard deviation for Method 2 is 11.73. 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 3:

Find the point estimate for the true difference between the population means.

Step 2 of 3:

Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Step 3 of 3:

Construct the 99%99% confidence interval. Round your answers to one decimal place.

Answer 1

1)

Point estimate = 53.9 - 88.9 = -35

2)

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(310.1121/214 + 137.5929/193)
sp = 1.4704

Given CI level is 0.99, hence α = 1 - 0.99 = 0.01                  
α/2 = 0.01/2 = 0.005,zc = z(α/2, df) = 2.576                  
                  
Margin of Error  
ME = tc * sp  
ME = 2.58 * 1.4704  
ME = 3.793632  
  

3)

                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (53.9 - 88.9 - 2.576 * 1.4704 , 53.9 - 88.9 - 2.576 * 1.4704                  
CI = (-38.8 , -31.2)