Question

the resting heart rates for 80 women aged 46–55 in a simple random sample are normally distributed, with a mean of 71 beats per minute and a standard deviation of 6 beats per minute. assuming a 90% confidence level (90% confidence level = z - score of 1.645), what is the margin of error for the population mean?
remember, the margin of error, me, can be determined using the formula $me = \frac{z \cdot s}{\sqrt{n}}$.
\\(\bigcirc\\) 0.66
\\(\bigcirc\\) 1.10
\\(\bigcirc\\) 1.31
\\(\bigcirc\\) 1.73

Explanation:

Step1: Identify given values

We have \( z = 1.645 \), \( \sigma = 6 \), and \( n = 80 \).

Step2: Substitute into the formula

The formula for margin of error is \( ME=\frac{z\cdot\sigma}{\sqrt{n}} \). Substitute the values: \( ME=\frac{1.645\times6}{\sqrt{80}} \).

Step3: Calculate the denominator

First, calculate \( \sqrt{80}\approx8.9443 \).

Step4: Calculate the numerator

Calculate \( 1.645\times6 = 9.87 \).

Step5: Divide numerator by denominator

Now, divide \( 9.87 \) by \( 8.9443 \): \( \frac{9.87}{8.9443}\approx1.10 \).

Answer:

1.10