To determine the margin of error (ME) for the population mean, we'll follow these steps and use the given formula: [tex]\(ME = \frac{z \cdot s}{\sqrt{n}}\)[/tex].
Let's break it down step-by-step:
1. Identify the given values:
- Sample size (n): [tex]\(80\)[/tex]
- Sample mean ( [tex]\(\bar{x}\)[/tex]): 71 beats per minute (though not required for the margin of error calculation)
- Sample standard deviation (s): [tex]\(6\)[/tex]
- [tex]\(z\)[/tex]-score for a 90% confidence level: [tex]\(1.645\)[/tex]
2. Plug these values into the formula:
[tex]\[
ME = \frac{z \cdot s}{\sqrt{n}}
\][/tex]
Substitute the given values:
[tex]\[
ME = \frac{1.645 \cdot 6}{\sqrt{80}}
\][/tex]
3. Calculate the denominator (the square root of the sample size):
[tex]\[
\sqrt{80} \approx 8.944
\][/tex]
4. Calculate the numerator (the product of [tex]\(z\)[/tex]-score and the standard deviation):
[tex]\[
1.645 \cdot 6 = 9.87
\][/tex]
5. Perform the division:
[tex]\[
ME = \frac{9.87}{8.944} \approx 1.1035
\][/tex]
So, rounding to two decimal places, the margin of error is approximately:
[tex]\[
ME \approx 1.10
\][/tex]
The answer is therefore [tex]\(1.10\)[/tex].
