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 mc014-1.jpg.

Answer 1

ME = 1.10.

Explanation:
The margin of error, ME, is given by the formula z*(σ/√n).

Since we want a 90% confidence interval, z=1.645.

σ=6, and n=80, giving us ME=1.645*(6/√80). This comes out to 1.10.

Answer 2

Answer:

The margin of error is $1.103$

Step-by-step explanation:

The margin of error represents the maximum expected difference between the true parameter and the estimated parameter.

Mathematically, it is calculated as,

$=z\cdot \dfrac{\sigma}{\sqrt{n}}$

where,

z = z score of the confidence interval,

σ = standard deviation,

n = sample size.

Putting the values,

$=1.645\times \dfrac{6}{\sqrt{80}}$

$=1.103$