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Certainly! Let's go through step-by-step to determine how large the sample must be in order for the length of the confidence interval to not exceed [tex]\(0.25 \sigma\)[/tex].
1. Understanding the Confidence Interval:
When estimating a population mean [tex]\(\mu\)[/tex] from a sample from a normal distribution [tex]\(N(\mu, \sigma^2)\)[/tex] with a known standard deviation [tex]\(\sigma\)[/tex], the confidence interval (CI) for [tex]\(\mu\)[/tex] at a confidence level of 95% can be expressed as:
[tex]\[ \bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \][/tex]
where:
- [tex]\(\bar{x}\)[/tex] is the sample mean
- [tex]\(z_{\alpha/2}\)[/tex] is the critical value from the standard normal distribution (for 95% confidence, [tex]\(z_{\alpha/2} \approx 1.96\)[/tex])
- [tex]\(\sigma\)[/tex] is the population standard deviation
- [tex]\(n\)[/tex] is the sample size
2. Length of the Confidence Interval:
The total length of the confidence interval is twice the margin of error, which is [tex]\(2 \cdot z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right)\)[/tex].
Given that this length must not exceed [tex]\(0.25 \sigma\)[/tex], we can set up the inequality:
[tex]\[ 2 \cdot z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \le 0.25 \sigma \][/tex]
3. Solving for the Sample Size [tex]\(n\)[/tex]:
First, divide both sides of the inequality by [tex]\(\sigma\)[/tex] to simplify:
[tex]\[ 2 \cdot z_{\alpha/2} \left(\frac{1}{\sqrt{n}}\right) \le 0.25 \][/tex]
Next, divide both sides by 2:
[tex]\[ z_{\alpha/2} \left(\frac{1}{\sqrt{n}}\right) \le 0.125 \][/tex]
Now, isolate [tex]\(\sqrt{n}\)[/tex] by dividing both sides by [tex]\(z_{\alpha/2} = 1.96\)[/tex]:
[tex]\[ \frac{1}{\sqrt{n}} \le \frac{0.125}{1.96} \][/tex]
Calculate the right-hand side:
[tex]\[ \frac{0.125}{1.96} \approx 0.06378 \][/tex]
Now, take the reciprocal of both sides:
[tex]\[ \sqrt{n} \ge \frac{1}{0.06378} \approx 15.68 \][/tex]
Finally, square both sides to solve for [tex]\(n\)[/tex]:
[tex]\[ n \ge (15.68)^2 \approx 245.86 \][/tex]
Thus, the minimum sample size [tex]\(n\)[/tex] required for the confidence interval length to be no more than [tex]\(0.25 \sigma\)[/tex] is approximately:
[tex]\[ n \approx 246 \][/tex]
So, [tex]\(n\)[/tex] should be at least 246.
1. Understanding the Confidence Interval:
When estimating a population mean [tex]\(\mu\)[/tex] from a sample from a normal distribution [tex]\(N(\mu, \sigma^2)\)[/tex] with a known standard deviation [tex]\(\sigma\)[/tex], the confidence interval (CI) for [tex]\(\mu\)[/tex] at a confidence level of 95% can be expressed as:
[tex]\[ \bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \][/tex]
where:
- [tex]\(\bar{x}\)[/tex] is the sample mean
- [tex]\(z_{\alpha/2}\)[/tex] is the critical value from the standard normal distribution (for 95% confidence, [tex]\(z_{\alpha/2} \approx 1.96\)[/tex])
- [tex]\(\sigma\)[/tex] is the population standard deviation
- [tex]\(n\)[/tex] is the sample size
2. Length of the Confidence Interval:
The total length of the confidence interval is twice the margin of error, which is [tex]\(2 \cdot z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right)\)[/tex].
Given that this length must not exceed [tex]\(0.25 \sigma\)[/tex], we can set up the inequality:
[tex]\[ 2 \cdot z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \le 0.25 \sigma \][/tex]
3. Solving for the Sample Size [tex]\(n\)[/tex]:
First, divide both sides of the inequality by [tex]\(\sigma\)[/tex] to simplify:
[tex]\[ 2 \cdot z_{\alpha/2} \left(\frac{1}{\sqrt{n}}\right) \le 0.25 \][/tex]
Next, divide both sides by 2:
[tex]\[ z_{\alpha/2} \left(\frac{1}{\sqrt{n}}\right) \le 0.125 \][/tex]
Now, isolate [tex]\(\sqrt{n}\)[/tex] by dividing both sides by [tex]\(z_{\alpha/2} = 1.96\)[/tex]:
[tex]\[ \frac{1}{\sqrt{n}} \le \frac{0.125}{1.96} \][/tex]
Calculate the right-hand side:
[tex]\[ \frac{0.125}{1.96} \approx 0.06378 \][/tex]
Now, take the reciprocal of both sides:
[tex]\[ \sqrt{n} \ge \frac{1}{0.06378} \approx 15.68 \][/tex]
Finally, square both sides to solve for [tex]\(n\)[/tex]:
[tex]\[ n \ge (15.68)^2 \approx 245.86 \][/tex]
Thus, the minimum sample size [tex]\(n\)[/tex] required for the confidence interval length to be no more than [tex]\(0.25 \sigma\)[/tex] is approximately:
[tex]\[ n \approx 246 \][/tex]
So, [tex]\(n\)[/tex] should be at least 246.
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