Question

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 students using Method 1 produces a testing average of 61. A sample of 156 students using Method 2 produces a testing average of 64.6. Assume the standard deviation is known to be 18.53 for Method 1 and 13.43 for Method 2. 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 2: Find the critical value that should be used in constructing the confidence interval.

Answer 1

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Explanation:

The (1 - α)% confidence interval for the difference between population means is:

`CI=(\bar x_(1)-\bar x_(2))\pm z_(\alpha/2)* \sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}`

The information provided is as follows:

`n_(1)= 138\\n_(2)=156\\\bar x_(1)=61\\\bar x_(2)=64.6\\\sigma_(1)=18.53\\\sigma_(2)=13.43`

The critical value of z for 98% confidence level is,

`z_(\alpha/2)=z_(0.02/2)=2.326`

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

`CI=(\bar x_(1)-\bar x_(2))\pm z_(\alpha/2)* \sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}`

`=(61-64.6)\pm 2.326*\sqrt{((18.53)^(2))/(138)+((13.43)^(2))/(156)}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)`

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).