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)**

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