To determine the required sample size for each location, we use the formula for the margin of error in estimating the difference between two means:

$ME = z_{\alpha/2} \times \sqrt{\frac{2\sigma^2}{n}}$

where:

• $ME$ is the margin of error (0.1 pH in this case),
• $z_{\alpha/2}$ is the critical z-score corresponding to a 90% confidence level (which is 1.645),
• $\sigma^2$ is the variance (given as 0.35),
• $n$ is the sample size per group.

Step 1: Solve for $n$

$0.1 = 1.645 \times \sqrt{\frac{2(0.35)}{n}}$

Step 2: Square both sides

$(0.1)^2 = (1.645)^2 \times \frac{2(0.35)}{n}$

$0.01 = 2.706 \times \frac{0.7}{n}$

$0.01 = \frac{1.8942}{n}$

Step 3: Solve for $n$

$n = \frac{1.8942}{0.01} = 189.42$

Since we always round up to the next whole number:

$n = 190$

Thus, the required sample size for each location is 190 rainfalls.

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Would you like a breakdown of any step? Here are some related questions:

1. How do we determine the critical z-score for different confidence levels?
2. What happens if the variances at the two locations are different?
3. How does increasing the confidence level affect the sample size?
4. Why do we round up the sample size instead of rounding to the nearest integer?
5. What are the implications of assuming equal variance in real-world studies?

Tip: If the confidence level were 95% instead of 90%, you would need a larger sample size because a higher confidence level requires a wider margin.