MATH SOLVE

3 months ago

Q:
# The resting heart rates for 80 women aged 46β55 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.

Accepted Solution

A:

The margin of error for the population mean for the 90% confidence level is 1.10.What is the margin of error?The margin of error is a statistic that expresses how much random sampling error there is in a survey's results. The wider the margin of error, the less confident one should be that a poll result reflects the outcome of a population-wide survey.It is given by the formula:[tex]MOE_y=Z_y\times \sqrt{\dfrac{\sigma^2}{n} }[/tex]As the mean is given to us is 71 beats and the standard deviation is 6 beats, therefore, the margin of error for 90% confidence level can be written as,[tex]MOE_y=Z_y\times \sqrt{\dfrac{\sigma^2}{n} }[/tex][tex]MOE_y=1.645\times \sqrt{\dfrac{\ 6^2}{80} }[/tex][tex]MOE_y= 1.645\times \sqrt{\dfrac{36}{80} }[/tex][tex]MOE_y= 1.10[/tex]Thus the margin of error for the population mean for the 90% confidence level is 1.10.To know more about Z- score follow