To find the appropriate values of $n_1$ and $n_2$ (where $n_1 = n_2 = n$), we can use the formula for the required sample size when estimating the difference in two means:

$n = \left( \frac{Z \cdot \sqrt{\sigma_1^2 + \sigma_2^2}}{E} \right)^2$

where:

• $Z$ is the critical value corresponding to the desired confidence level,
• $\sigma_1$ and $\sigma_2$ are the population standard deviations,
• $E$ is the maximum allowable sampling error.

Part (a)

For a sampling error of 3.1 with 95% confidence:

• $E = 3.1$
• $\sigma_1 = 10$
• $\sigma_2 = 20$
• For a 95% confidence level, $Z = 1.96$

Let's calculate:

$n = \left( \frac{1.96 \cdot \sqrt{10^2 + 20^2}}{3.1} \right)^2$

1. Calculate the standard deviation term: $\sqrt{10^2 + 20^2} = \sqrt{100 + 400} = \sqrt{500} \approx 22.36$

2. Substitute values and calculate $n$: $n = \left( \frac{1.96 \cdot 22.36}{3.1} \right)^2 = \left( \frac{43.8256}{3.1} \right)^2 \approx (14.14)^2 \approx 200$

Thus, for part (a), the sample size needed is approximately 200.

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Part (b)

For a sampling error of 6 with 99% confidence:

• $E = 6$
• The range of each population is 48. To estimate the standard deviations $\sigma_1$ and $\sigma_2$, we can use the approximation $\sigma \approx \frac{\text{Range}}{4}$. So, $\sigma_1 \approx \sigma_2 \approx \frac{48}{4} = 12$.
• For a 99% confidence level, $Z = 2.576$.

Now calculate:

$n = \left( \frac{2.576 \cdot \sqrt{12^2 + 12^2}}{6} \right)^2$

1. Calculate the standard deviation term: $\sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} \approx 16.97$

2. Substitute values and calculate $n$: $n = \left( \frac{2.576 \cdot 16.97}{6} \right)^2 = \left( \frac{43.7192}{6} \right)^2 \approx (7.2865)^2 \approx 53.1$

Rounding up, the sample size needed for part (b) is approximately 54.

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Part (c)

For a 90% confidence interval of width 1.7:

• Desired width of the interval, $2E = 1.7$, so $E = \frac{1.7}{2} = 0.85$.
• $\sigma_1^2 \approx 5.1$ and $\sigma_2^2 \approx 7.5$, thus $\sigma_1 = \sqrt{5.1} \approx 2.26$ and $\sigma_2 = \sqrt{7.5} \approx 2.74$.
• For a 90% confidence level, $Z = 1.645$.

Now calculate:

$n = \left( \frac{1.645 \cdot \sqrt{2.26^2 + 2.74^2}}{0.85} \right)^2$

1. Calculate the standard deviation term: $\sqrt{2.26^2 + 2.74^2} = \sqrt{5.1076 + 7.5076} = \sqrt{12.6152} \approx 3.55$

2. Substitute values and calculate $n$: $n = \left( \frac{1.645 \cdot 3.55}{0.85} \right)^2 = \left( \frac{5.83675}{0.85} \right)^2 \approx (6.8685)^2 \approx 47.2$

Rounding up, the sample size needed for part (c) is approximately 48.

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Summary

• (a): Sample size needed = 200
• (b): Sample size needed = 54
• (c): Sample size needed = 48

Would you like more explanation on these steps, or have any further questions?

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Additional Questions:

1. How does the critical value $Z$ change with different confidence levels?
2. Why do we use the approximation $\sigma \approx \frac{\text{Range}}{4}$ for estimating the standard deviation?
3. What role does the sampling error $E$ play in determining sample size?
4. How would these calculations change if $\sigma_1 \neq \sigma_2$?
5. What if a 99% confidence interval was needed for part (c)?

Tip:

The critical value $Z$ directly impacts sample size: higher confidence levels require larger $Z$ values, which in turn increase $n$ for the same error margin $E$.