Question

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 206206 students using Method 1 produces a testing average of 82.9. A sample of 244 students using Method 2 produces a testing average of 62.2. Assume that the population standard deviation for Method 1 is 10.23, while the population standard deviation for Method 2 is 11.35. Determine the 95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

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 95% confidence interval. Round to 2 decimals.

Answer 1

Answer :

Given data is :

Sample size

Mean =

Standard deviation =

Confidence interval is at 95%

i.e z value at 95% is 1.96

therefore,

Step 2 :

Margin of error ME =

=

=

=

=

=

=

Margin of error ME = 1.995

Step 3 :

Now,confidence interval at 95% is :

=(82.9 - 62.2) +/- 1.995

= 20.4 +/- 1.995

= (20.4 - 1.995 , 20.4 + 1.995)

= (18.405 , 22.395)

confidence interval at 95% = (18.405 , 22.395)