Respuesta :
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
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.
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The margin of error is given by:
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
Considering the standard deviation and the sample size, we have that [tex]s = 6, n = 80[/tex].
The critical value is t = 1.645.
Hence:
[tex]M = 1.645\frac{6}{\sqrt{80}} = 1.1[/tex]
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
