A researcher compares the effectiveness of two different ins

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 207 students using Method 1 produces a testing average of 84.4. A sample of 193 students using Method 2 produces a testing average of 77.4. Assume that the population standard deviation for Method 1 is 15.57, while the population standard deviation for Method 2 is 17.84. Determine the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

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

Step 2: 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: Construct the 98% confidence interval. Round your answers to one decimal place. Lower endpoint? Upper endpoint?

Solution

CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x1)=84.4
Standard deviation( sd1 )=15.57
Sample Size(n1)=207
Mean(x2)=77.4
Standard deviation( sd2 )=17.84
Sample Size(n12=193
CI = [ ( 84.4-77.4) ±Z a/2 * Sqrt( 242.4249/207+318.2656/193)]
= [ (7) ± Z a/2 * Sqrt( 2.8202) ]
= [ (7) ± 2.326 * Sqrt( 2.8202) ]
= [3.0939 , 10.9061]

[ANSWERS]
a.
Point of estimate = ( 84.4-77.4) = 7

b.
Margin Of Error = Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
= 2.326 * Sqrt( 242.4249/207+318.2656/193)
= 3.9061
c.[3.0939 , 10.9061]