Assume that a sample is used to estimate a population mean $\mu$. Find the margin of error that corresponds to a sample of size 18 with a mean of 44.4 and a standard deviation of 6.2 at a confidence level of $80\%$.



Round to one decimal place because the sample statistics are presented with this accuracy.

$E=$

Verified Answer :

  • Identify the degrees of freedom: \(df = n - 1 = 17\).
  • Determine the critical t-value for an \(80\%\) confidence level: \(t_{0.10, 17} \approx 1.333\).
  • Calculate the margin of error: \(E = t_{0.10, 17} \cdot \frac{s}{\sqrt{n}} = 1.333 \cdot \frac{6.2}{\sqrt{18}} \approx 1.9485\).
  • Round the result to one decimal place: \(\boxed{1.9}\).

Explanation

  1. Analysis and parameters identification

To find the margin of error \(E\) for the population mean \(\mu\) when the population standard deviation \(\sigma\) is unknown, we use the t-distribution.
The given data is:

  • Sample size: \(n = 18\)
  • Sample mean: \(\bar{x} = 44.4\)
  • Sample standard deviation: \(s = 6.2\)
  • Confidence level: \(80\%\)

First, we determine the degrees of freedom (\(df\)), which is calculated as \(n - 1\):
\(df = 18 - 1 = 17\).

Next, we identify the critical t-value (\(t_{\alpha/2}\)) for the \(80\%\) confidence level. With \(\alpha = 1 - 0.80 = 0.20\), we look for the t-value corresponding to the upper tail probability of \(\alpha/2 = 0.10\) with \(17\) degrees of freedom. The critical t-value for \(df = 17\) and a tail area of \(0.10\) is approximately \(1.333\).

  1. Calculation of the margin of error

The formula for the margin of error \(E\) is:
\[E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]
Using the values identified:

  • \(t_{\alpha/2} \approx 1.333\)
  • \(s = 6.2\)
  • \(\sqrt{n} = \sqrt{18} \approx 4.2426\)

Now, we substitute these values into the formula:
\[E = 1.333 \cdot \frac{6.2}{\sqrt{18}}\]
\[E = 1.333 \cdot \frac{6.2}{4.2426}\]
\[E \approx 1.333 \cdot 1.4614 \approx 1.9485\]

  1. Rounding and final conclusion

The calculated margin of error is approximately \(1.9485\). We are asked to round the result to one decimal place, consistent with the precision of the sample statistics provided.
Rounding \(1.9485\) to one decimal place gives \(1.9\).

Therefore, the margin of error is \(\boxed{1.9}\).

Examples

In a real-world scenario, a researcher might measure the heart rate of 18 participants to estimate the average heart rate of an entire population. By calculating the margin of error, the researcher can establish a confidence interval—a range of values within which the true population mean is likely to lie—thereby providing a clear measure of the uncertainty in their estimate. This is essential for ensuring that research findings are reliable and statistically sound.