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?
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
Using the t-distribution, it is found that the margin of error for the population mean is of 1.1 beats per minute.
We have the standard deviation for the sample, hence the t-distribution is used to solve this question.
The confidence interval is:
\(\overline{x} \pm t\frac{s}{\sqrt{n}}\)
In which:
The margin of error is given by:
\(M = t\frac{s}{\sqrt{n}}\)
Considering the standard deviation and the sample size, we have that \(s = 6, n = 80\).
The critical value is t = 1.645.
Hence:
\(M = 1.645\frac{6}{\sqrt{80}} = 1.1\)
The margin of error for the population mean is of 1.1 beats per minute.
You can learn more about the t-distribution at https://brainly.com/question/25256953