- The survey sample size is \(n=27\) with a standard deviation of \(s=\$7\).
- Using the t-distribution with \(df=26\) and a confidence level of \(80\%\), the critical t-value is \(t_{0.10} \approx 1.315\).
- The margin of error is calculated as \(E = t_{0.10} \times \frac{s}{\sqrt{n}} \approx 1.315 \times \frac{7}{\sqrt{27}}\).
- Rounding to two decimal places, the margin of error is \(\boxed{\$1.77}\).
Explanation
- Analyze the problem and given data
The survey provides data from a sample size of \(n = 27\) people. The distribution of the spending is roughly bell-shaped, and since the population standard deviation \(\sigma\) is unknown, we use the t-distribution to estimate the margin of error. The sample standard deviation is \(s = \$7\). Our goal is to find the margin of error \(E\) for an \(80\%\) confidence level.
- Determine the critical t-value and formula components
The margin of error \(E\) is calculated using the formula:
\[E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]
where \(t_{\alpha/2}\) is the critical t-value corresponding to the desired confidence level and degrees of freedom \(df = n - 1\). With a confidence level of \(80\%\), the significance level \(\alpha\) is \(1 - 0.80 = 0.20\). Thus, we need the critical value for \(\alpha/2 = 0.10\). The degrees of freedom are \(df = 27 - 1 = 26\). Using a t-distribution table or statistical software for \(df = 26\) and an area of \(0.10\) in the right tail, we find the critical t-value \(t_{0.10} \approx 1.315\).
- Calculate the margin of error
Now, we substitute the values into the margin of error formula:
\[E = 1.315 \times \frac{7}{\sqrt{27}}\]
First, calculate the standard error: \(\frac{7}{\sqrt{27}} \approx \frac{7}{5.196} \approx 1.347\).
Then, multiply by the critical value: \(E \approx 1.315 \times 1.347 \approx 1.7715\).
Rounding the margin of error to 2 decimal places, we get \(E \approx 1.77\).
Examples
Margin of error is widely used in market research and public opinion polling to understand the precision of a survey estimate. For instance, if a company surveys customers to see how much they spend on gifts, the margin of error tells them how much the true average spending might deviate from their sample mean at a specific confidence level. This helps businesses make more informed decisions based on the reliability of their data. For example, if the sample mean is \(\bar{x} = \$42\) with a margin of error \(E = \$1.77\), the company can be \(80\%\) confident that the true population average spend is between \(\$40.23\) and \(\$43.77\).