Question

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 175 students using Method 1 produces a testing average of 89.6. A sample of 177 students using Method 2 produces a testing average of 68.8. Assume the standard deviation is known to be 17.56 for Method 1 and 5.07 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.

Step 2 of 2:

Construct the 98% confidence interval. Round your answers to one decimal place.

Lower endpoint: Upper endpoint:

Answer 1

Step 1

Z(α/2) = Z (0.02 /2) = 2.326

Step 2

Confidence interval :-

Z(α/2) = Z (0.02 /2) = 2.326

Lower endpoint =
Lower endpoint = 17.5877 ≈ 17.6
Upper endpoint =
Upper endpoint = 24.0123 ≈ 24.0
98% Confidence interval is ( 17.6 , 24.0 )

Answer 2

Step 1:

We need to find the critical value for a 98% confidence interval with 175-1=174 degrees of freedom (df) and a two-tailed test. We can use a t-distribution since the population standard deviations are unknown.

Using a t-distribution table or calculator, the critical value with 174 df and a 98% confidence level is approximately 2.333.

Step 2:

To construct the confidence interval, we can use the formula:

CI = (x̄1 - x̄2) ± t*(SE)

where x̄1 and x̄2 are the sample means for Method 1 and Method 2, respectively, t is the critical value we found in Step 1, and SE is the standard error of the difference between the means.

The standard error of the difference between the means is calculated as:

SE = sqrt((s1^2 / n1) + (s2^2 / n2))

where s1 and s2 are the known population standard deviations, n1 and n2 are the sample sizes for Method 1 and Method 2, respectively.

Plugging in the values given in the problem, we get:

SE = sqrt((17.56^2 / 175) + (5.07^2 / 177)) = 1.920

Now we can plug in all the values to get the confidence interval:

CI = (89.6 - 68.8) ± 2.333*(1.920) = 20.8 ± 4.476

Rounding to one decimal place, we get:

Lower endpoint: 16.3 Upper endpoint: 25.3

Therefore, we are 98% confident that the true difference between the testing averages for students using Method 1 and students using Method 2 is between 16.3 and 25.3.