Question

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 54 students using Method 1 produces a testing average of 51.7. A sample of 90 students using Method 2 produces a testing average of 56.8. Assume that the population standard deviation for Method 1 is 7.35, while the population standard deviation for Method 2 is 16.72. Determine the 80% 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 80% confidence interval. Round your answers to one decimal place.

Answer 1

1)

the point estimate for the true difference between the population means = 51.7 - 56.8 = -5.10

2)

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(54.0225/54 + 279.5584/90)
sp = 2.0265

Given CI level is 0.8, hence α = 1 - 0.8 = 0.2                  
α/2 = 0.2/2 = 0.1, zc = z(α/2, df) = 1.28                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 1.28 * 2.0265                  
ME = 2.593920

3)

                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (51.7 - 56.8 - 1.28 * 2.0265 , 51.7 - 56.8 - 1.28 * 2.0265                  
CI = (-7.7 , -2.5)