A researcher compares the effectiveness of two different instructional methods for teaching physiology.  A sample of 297297 students using Method 1 produces a testing average of 50.250.2.  A sample of 298298 students using Method 2 produces a testing average of 70.870.8.  Assume the standard deviation is known to be 11.0511.05 for Method 1 and 7.277.27 for Method 2.  Determine the 98%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.

To find the critical value for constructing a confidence interval at a 98% confidence level, we'll use the \( z \)-score corresponding to 98% confidence. Since we are working with two independent samples and the standard deviations are known, we use the normal distribution \( z \)-score for this calculation.

### Step-by-Step Solution
1. **Identify the Confidence Level**: We have a 98% confidence level.

2. **Determine the Tail Probability**: For a 98% confidence interval, there is a 2% (or 0.02) total probability in the two tails of the normal distribution (1 - 0.98 = 0.02). Since the distribution is symmetric, each tail has 1% (or 0.01) of the probability.

3. **Find the Critical Value \( z_{\alpha/2} \)**: 
   - We need the \( z \)-score that leaves 1% in the upper tail.
   - This \( z \)-score can be found from a standard normal distribution table, or through a calculator. 

   For a 98% confidence level, the critical \( z \)-value (denoted \( z_{\alpha/2} \)) is approximately:
   \[
   z_{\alpha/2} \approx 2.33
   \]

### Answer
The critical value to use for constructing a 98% confidence interval is:
\[
z_{\alpha/2} = 2.33
\]

Would you like to proceed with the next step of calculating the confidence interval?

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Here are some follow-up questions to deepen understanding:
1. Why is a \( z \)-score used instead of a \( t \)-score in this scenario?
2. How would the critical value change if we had a 95% confidence level instead?
3. What would the interpretation of a 98% confidence interval be in this context?
4. How does sample size influence the width of the confidence interval?
5. Why do we divide the tail probability by two for a two-tailed confidence interval?

**Tip**: For higher confidence levels, the critical value increases, making confidence intervals wider.

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 297 students using Method 1 produces a testing average of 50.2. A sample of 298 students using Method 2 produces a testing average of 70.8. Assume the standard deviation is known to be 11.05 for Method 1 and 7.27 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.

Determine the 98% confidence interval for the difference between two instructional methods in teaching physiology. This solution covers key statistical concepts like the normal distribution and confidence intervals, including a critical z-score value calculation. Suitable for undergraduate students studying statistics.

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