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

Answer:

(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 $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$, i.e.,

$\sqrt{\frac{(17.56)^{2}}{175}+\frac{(5.07)^{2}}{177}}$ = 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