Question

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 189 students using Method 1 produces a testing average of 71.9. A sample of 141 students using Method 2 produces a testing average of 78.3. Assume that the population standard deviation for Method 1 is 11.07, while the population standard deviation for Method 2 is 11.9. 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.

Answer #1

Step 1 of 3: Find the point estimate for the true difference between the population means.

point estimate=x1bar-x2bar

= 71.9- 78.3

=-6.4

-6.4

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.

z crit for 98% in excel

=NORM.S.INV(0.99)

=2.326347874

margin of error of a confidence interval for the difference between the two population means.

=zcrit*sqrt(sigma1^2/n1+sigma2^2/n2)

=2.326347874*sqrt((11.07^2/189)+(11.9^2/141))

= 2.990703

2.990703

Step 3 of 3: Construct the 98% confidence interval. Round your answers to one decimal place

point estimate-margin of error,point estimate+margin of error

-6.4-2.990703,-6.4+2.990703

-9.390703, -3.409297

-9.390703<mu1-mu2< -3.409297

(-9.4,-3.4)

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