Question

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 215215 students using Method 1 produces a testing average of 55.555.5. A sample of 242242 students using Method 2 produces a testing average of 64.264.2. Assume that the population standard deviation for Method 1 is 7.957.95, while the population standard deviation for Method 2 is 18.2118.21. Determine the 90%90% 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.

Answer 1

Let's write the given information:

n₁ = sample size for method 1 = 215

= sample mean for method 1 = 55.5

population standard deviation for method 1 = 7.95

n₂ = sample size for method 2 = 242

= sample mean for method 2 = 64.2

population standard deviation for method 2 = 18.21

c = confidence level = 0.90

= level of significance = 1 - c = 1 - 0.90 = 0.90

step 1:  The point estimate for the true difference between the population means is as follows.

Point estimate =

step 2:

Let's find margin of error ( E ).

Formula of margin of error ( E ) is as follows:

Here is the critical Z value.

For 90% confidence level = 1.645

Step 3:

Lower limit : = Point estimate - E = -8.7 + 2.12 = -10.82

Upper limit : = Point estimate + E = -8.7 + 2.12 = -6.58

Therefore 90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is ( -10.82 , -6.58 )