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 m e = startfraction z times s over startroot n endroot endfraction. 0.66 1.10 1.31 1.73

Respuesta :

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]\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\frac{\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]\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\frac{\sigma^{2}}{n}}\\\\\text{MOE}_{\gamma}=1.645 \times \sqrt{\dfrac{6^{2}}{80}}\\\\\text{MOE}_{\gamma}=1.645 \times \sqrt{\dfrac{36}{80}}\\\\\text{MOE}_{\gamma}=1.1034\approx 1.10[/tex]

Hence, the margin of error for the population mean for the 90% confidence level is 1.10.

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