A researcher compares the effectiveness of two different instructional methods f
ID: 3359070 • Letter: A
Question
A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 225 students using Method 1 produces a testing average of 68.2. A sample of 242 students using Method 2 produces a testing average of 66.2. Assume that the population standard deviation for Method 1 is 5.66, while the population standard deviation for Method 2 is 10.06. Determine the 98% 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 98% confidence interval. Round your answers to one decimal place.
Explanation / Answer
The statistical software output for this problem is:
Two sample Z summary confidence interval:
1 : Mean of population 1 (Std. dev. = 5.66)
2 : Mean of population 2 (Std. dev. = 10.06)
1 - 2 : Difference between two means
98% confidence interval results:
Hence,
1) Point estimate = 2
2) Margin of error = (3.7417762 - 0.25822381)/2 = 1.741776
3) 98% confidence interval: (0.3, 3.7)
Difference n1 n2 Sample mean Std. err. L. limit U. limit 1 - 2 225 242 2 0.748717 0.25822381 3.7417762Related Questions
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