Question

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 169 students using method 1 produces a testing average of 85.2. A sample of 127 students using method 2 produces a testing average of 56.4. Assume that the population standard deviation for method 1 is 15.27, while the population standard deviation for method 2 is 15.46. Determine the 98th percentile for the difference in the means of the two methods.

Answer 1

To determine the 98th percentile for the difference in means, calculate the standard error of the difference using the formula SE_d = sqrt(SD_1^2/n_1 + SD_2^2/n_2), find the z-score corresponding to the 98th percentile, and multiply the standard error by the z-score to get the margin of error. Add the margin of error to the difference between the sample means to find the 98th percentile.

Step-by-step explanation:

To determine the 98th percentile for the difference in means, we need to calculate the standard error of the difference. The formula for the standard error of the difference is:

SEd = sqrt(SD12/n1 + SD22/n2)

where SD1 and SD2 are the population standard deviations and n1 and n2 are the sample sizes. Plug in the given values:

SEd = sqrt(15.272/169 + 15.462/127) = 1.3225

To find the 98th percentile, we need to determine the z-score corresponding to that percentile. Look up the z-score in a standard normal distribution table or use a calculator:

Z98 = 2.0537

Finally, calculate the margin of error by multiplying the standard error by the z-score:

ME = 2.0537 * 1.3225 = 2.7175

Therefore, the 98th percentile for the difference in means is the difference between the sample means plus the margin of error:

98th percentile = (85.2 - 56.4) + 2.7175 = 31.5175