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.

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

Step-by-step 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.

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

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

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

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

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

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

(-1,-0.5)

step-by-step explanation: