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Margin of Error Calculation for a Confidence Interval

A sample of size \(n=64\) has sample mean \(\bar{x}=52.4\). The population standard deviation is known to be \(\sigma=10\). For 95% confidence, what is the margin of error calculation and the resulting confidence interval for \(\mu\)?

Subject: Statistics Chapter: Estimation of the Mean and Proportion Topic: Point and Interval Estimates Answer included
margin of error calculation margin of error confidence interval z critical value standard error estimation of mean sigma known sample size
Accepted answer Answer included

Meaning of margin of error calculation

In estimation, a margin of error calculation produces a nonnegative number \(E\) such that a confidence interval has the form \[ \text{estimate} \pm E. \] For a population mean with known \(\sigma\), the interval is centered at \(\bar{x}\) and the margin of error depends on a critical value and the standard error.

Known-\(\sigma\) mean (z interval)

\[ E = z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}, \qquad \text{CI for }\mu:\ \bar{x} \pm E. \]

Here \(\alpha=1-\text{confidence level}\), and \(z_{\alpha/2}\) is the standard normal critical value with upper-tail area \(\alpha/2\).

Step-by-step solution for the given data

Given \(n=64\), \(\bar{x}=52.4\), \(\sigma=10\), and 95% confidence: \[ \alpha = 1-0.95 = 0.05, \quad \alpha/2=0.025, \quad z_{\alpha/2}=z_{0.025}=1.96. \]

1) Compute the standard error

\[ \frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{64}}=\frac{10}{8}=1.25. \]

2) Margin of error calculation

\[ E=z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}} =1.96\cdot 1.25 =2.45. \]

3) Confidence interval

\[ \bar{x}\pm E = 52.4 \pm 2.45. \] Lower endpoint: \[ 52.4 - 2.45 = 49.95. \] Upper endpoint: \[ 52.4 + 2.45 = 54.85. \]

Result

Margin of error: \(E=2.45\).
95% confidence interval for \(\mu\): \((49.95,\ 54.85)\).

Visualization: estimate \(\pm E\) on a number line

The center mark is \(\bar{x}=52.4\). The endpoints are \(\bar{x}-E=49.95\) and \(\bar{x}+E=54.85\). The bracket length corresponds to \(2E\).

49.95 52.4 54.85 95% CI: \(\bar{x}\pm E\) \(E=2.45\) \(E=2.45\)
The margin of error is the distance from the point estimate \(\bar{x}\) to either endpoint of the confidence interval.

Reference: common critical values for margin of error calculation

For a two-sided z-based interval (known \(\sigma\)), \(z_{\alpha/2}\) depends only on the confidence level.

Confidence level \(\alpha\) \(\alpha/2\) \(z_{\alpha/2}\)
90% 0.10 0.05 1.645
95% 0.05 0.025 1.96
99% 0.01 0.005 2.576

Extensions frequently needed in practice

Mean with \(\sigma\) unknown (t interval)

When \(\sigma\) is unknown, replace \(\sigma\) by sample standard deviation \(s\) and use a t critical value.

\[ E = t_{\alpha/2,\,n-1}\cdot \frac{s}{\sqrt{n}}, \qquad \text{CI for }\mu:\ \bar{x}\pm E. \]

Population proportion (large-sample z interval)

For \(\hat{p}\) with sufficiently large counts, the margin of error uses the standard error of \(\hat{p}\).

\[ E = z_{\alpha/2}\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \qquad \text{CI for }p:\ \hat{p}\pm E. \]

Sample size planning from a target margin of error

  • Mean (known \(\sigma\)): to achieve margin \(E\), \[ n \ge \left(\frac{z_{\alpha/2}\cdot \sigma}{E}\right)^2. \]
  • Proportion: to achieve margin \(E\), \[ n \ge \frac{z_{\alpha/2}^2\cdot \hat{p}(1-\hat{p})}{E^2}, \] and the conservative choice \(\hat{p}=0.5\) maximizes \(\hat{p}(1-\hat{p})\).
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