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.

Answer 1

To find the 98% confidence interval for the difference between the testing averages for students using Method 1 and Method 2, use the formula CI = (x1 - x2) ± z * sqrt((s1^2 / n1) + (s2^2 / n2)). Plug in the values to get the interval.

Step-by-step explanation:

To find the 98% confidence interval for the difference between the testing averages for students using Method 1 and Method 2, we can use the formula:

CI = (x1 - x2) ± z * sqrt((s1^2 / n1) + (s2^2 / n2))

where:

• CI is the confidence interval
• x1 and x2 are the sample means (89.6 and 68.8)
• s1 and s2 are the standard deviations (17.56 and 5.07)
• n1 and n2 are the sample sizes (175 and 177)
• z is the z-score corresponding to the desired confidence level (2.33 for a 98% confidence level)

Plugging in the values, we get:

CI = (89.6 - 68.8) ± 2.33 * sqrt((17.56^2 / 175) + (5.07^2 / 177))

Simplifying this will give you the 98% confidence interval for the true difference between the testing averages of the two methods.

Answer 2

(17.5874, 24.0126) is the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step-by-step explanation:

Let
`\mu_(1)-\mu_(2)` be the true difference between testing averages for students using Method 1 and students using Method 2. We have the sample sizes
`n_(1) = 175` and
`n_(2) = 177`, the unbiased point estimate for
`\mu_(1)-\mu_(2)` is
`\bar{x}_(1) - \bar{x}_(2)`, i.e., 89.6 - 68.8 = 20.8

The standard error is given by
`\sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}`, i.e.,

`\sqrt{((17.56)^(2))/(175)+((5.07)^(2))/(177)}` = 1.3810. Then, the endpoints for a 98% confidence interval for
`\mu_(1)-\mu_(2)` is given by

20.8-
`z_(0.02/2)`1.3810 and 20.8+
`z_(0.02/2)`1.3810, i.e.,

20.8-
`z_(0.01)`1.3810 and 20.8+
`z_(0.01)`1.3810 where
`z_(0.01)` is the the first quantile of the standard normal distribution, i.e., -2.3263, so, we have

20.8-(2.3263)(1.3810) and 20.8+(2.3263)(1.3810), i.e.,

17.5874 and 24.0126