Question

if n=17, \\(\bar{x}\\)=41, and s=9, find the margin of error at a 80% confidence level. give your answer to two decimal places.

Explanation:

Step1: Identify the distribution

Since the sample size \( n = 17 \) (small, \( n<30 \)) and population standard deviation is unknown, we use the t - distribution. The formula for the margin of error \( E \) for a confidence interval of the mean is \( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}} \), where \( t_{\alpha/2} \) is the critical value, \( s \) is the sample standard deviation, and \( n \) is the sample size.

First, we need to find the degrees of freedom \( df=n - 1=17 - 1 = 16 \). The confidence level is \( 80\% \), so the significance level \( \alpha=1 - 0.8=0.2 \), and \( \alpha/2 = 0.1 \).

Step2: Find the critical value \( t_{\alpha/2} \)

Using a t - table or a calculator with t - distribution functionality, for \( df = 16 \) and \( \alpha/2=0.1 \), we find that \( t_{0.1,16}\approx1.337 \) (we can look this up in the t - distribution table: the row for \( df = 16 \) and the column for one - tailed probability \( 0.1 \)).

Step3: Calculate the margin of error

We know that \( s = 9 \), \( n = 17 \), and \( t_{\alpha/2}=1.337 \). Substitute these values into the formula for the margin of error:

\( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}=1.337\times\frac{9}{\sqrt{17}} \)

First, calculate \( \sqrt{17}\approx4.123 \). Then \( \frac{9}{4.123}\approx2.183 \). Then multiply by \( 1.337 \): \( 1.337\times2.183\approx2.92 \)

Answer:

\( 2.92 \)