Question

A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 63 students using Method 1 produces a testing average of 52.2. A sample of 93 students using Method 2 produces a testing average of 72.3. Assume that the population standard deviation for Method 1 is 15.92, while the population standard deviation for Method 2 is 17.96. Determine the 95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Answer 1

The 95% confidence interval for the difference between means is (-25.5, -14.7).

Explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1 (Method 1), of size n1=63 has a mean of 52.2 and a standard deviation of 15.92.

The sample 2 (Method 2), of size n2=93 has a mean of 72.3 and a standard deviation of 17.96.

The difference between sample means is Md=-20.1.

`M_d=M_1-M_2=52.2-72.3=-20.1`

The estimated standard error of the difference between means is computed using the formula:

`s_(M_d)=\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}=\sqrt{(15.92^2)/(63)+(17.96^2)/(93)}\\\\\\s_(M_d)=√(4.023+3.468)=√(7.491)=2.74`

The critical t-value for a 95% confidence interval is t=1.975.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_(M_d)=1.975 \cdot 2.74=5.41`

Then, the lower and upper bounds of the confidence interval are:

`LL=M_d-t \cdot s_(M_d) = -20.1-5.41=-25.5\\\\UL=M_d+t \cdot s_(M_d) = -20.1+5.41=-14.7`

The 95% confidence interval for the difference between means is (-25.5, -14.7).