Question

question 10 of 13 step 2 of 2 a researcher compares the effectiveness of two different instructional methods for teaching anatomy. a sample of 158 students using method 1 produces a testing average of 67.5. a sample of 179 students using method 2 produces a testing average of 55.2. assume the standard deviation is known to be 17.81 for method 1 and 19.32 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 2 of 2: construct the 98% confidence interval. round your answers to one decimal place. answerhow to enter your answer (opens in new window) 1 point lower endpoint: upper endpoint:

Explanation:

Step1: Identify the given values

Let $\bar{x}_1 = 67.5$ (mean of Method 1), $n_1=158$ (sample - size of Method 1), $\sigma_1 = 17.81$ (standard - deviation of Method 1), $\bar{x}_2 = 55.2$ (mean of Method 2), $n_2 = 179$ (sample - size of Method 2), $\sigma_2=19.32$ (standard - deviation of Method 2). For a 98% confidence interval, the z - value $z_{\alpha/2}$ is 2.33.

Step2: Calculate the standard error

The formula for the standard error $SE$ of the difference between two means when the population standard - deviations are known is $SE=\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}$.
\[

$$\begin{align*} SE&=\sqrt{\frac{17.81^{2}}{158}+\frac{19.32^{2}}{179}}\\ &=\sqrt{\frac{317.1961}{158}+\frac{373.2624}{179}}\\ &=\sqrt{2.00757 + 2.08526}\\ &=\sqrt{4.09283}\\ &\approx2.023 \end{align*}$$

\]

Step3: Calculate the margin of error

The margin of error $ME = z_{\alpha/2}\times SE$.
$ME=2.33\times2.023\approx4.714$.

Step4: Calculate the lower and upper endpoints

The difference in sample means is $\bar{x}_1-\bar{x}_2=67.5 - 55.2 = 12.3$.
The lower endpoint $=(\bar{x}_1-\bar{x}_2)-ME=12.3-4.714\approx7.6$.
The upper endpoint $=(\bar{x}_1-\bar{x}_2)+ME=12.3 + 4.714\approx17.0$.

Answer:

Lower endpoint: 7.6
Upper endpoint: 17.0