the resting heart rates for 80 women aged 46 - 55 in a simple random sample are normally distributed, with a…

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 $me = \frac{zcdot s}{sqrt{n}}$. 0.66 1.10 1.31 1.73

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 $me = \frac{zcdot s}{sqrt{n}}$. 0.66 1.10 1.31 1.73

Answer

Explanation:

Step1: Identify values

$z = 1.645$, $s=6$, $n = 80$

Step2: Substitute into formula

$ME=\frac{z\cdot s}{\sqrt{n}}=\frac{1.645\times6}{\sqrt{80}}$

Step3: Calculate square - root

$\sqrt{80}\approx8.944$

Step4: Calculate numerator

$1.645\times6 = 9.87$

Step5: Calculate margin of error

$ME=\frac{9.87}{8.944}\approx1.10$

Answer:

1.10