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?

Answer 1

Given:
Confidence level = 90%
mean = 71 beats per minute
standard deviation = 6 beats per minute

margin of error = z * δ / √n

where : δ - population of the standard deviation, n is the sample size ; z is the appropriate z value.

90% confidence level = 1.645 in z-value

margin of error = 1.645 * (6/√80) = 1.645 * (6/8.94) = 1.645 *  0.671 = 1.104

Answer 2

Answer:

1.10/ answer B

Step-by-step explanation: