If $n=16, \bar{x}=31$, and $s =7$, find the margin of error at a $99 \%$ confidence level. Round to 2 decimal places.
$E=$
To determine the margin of error for the population mean when the population standard deviation is unknown, we use the t-distribution.
Given data:
First, we determine the degrees of freedom, which is \(df = n - 1 = 16 - 1 = 15\).
Next, for a \(99\%\) confidence level, the significance level \(\alpha\) is \(1 - 0.99 = 0.01\). We are looking for the critical value \(t_{\alpha/2, df} = t_{0.005, 15}\).
Based on the t-distribution table, the critical value \(t_{0.005, 15}\) is approximately \(2.947\).
The margin of error \(E\) is calculated using the formula:
\[E = t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}}\]
Substituting the known values into the equation:
Now, compute the margin of error:
\[E = 2.947 \cdot \frac{7}{4}\]
\[E = 2.947 \cdot 1.75\]
\[E = 5.15725\]
Rounding this result to 2 decimal places, we get \(E = 5.16\).
In medical research, if you want to estimate the average blood pressure reduction in a new patient group of size 16 with a 99% confidence interval, you use the t-distribution. By finding the margin of error, you can present a range within which the true population mean reduction likely lies, providing a more reliable understanding of the treatment's impact despite small sample sizes. For example, with a sample standard deviation of \(s = 7\), the calculated margin of error \(\approx 5.16\) indicates the precision of your estimate.