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There were 324 adults surveyed. Among the participants, the mean number of hours of sleep each night was 7. 5 and the standard deviation was 1. 6. The margin of error, assuming a 95% confidence level, is approximately. Round to the nearest hundredth.

Sagot :

The margin of error for 324 adults surveyed with 1.6 standard deviations is 0.1742.

What is the margin of error?

The margin of error can be defined as the amount of random sampling error in the results of a survey. It is given by the formula,

[tex]\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\frac{\sigma^{2}}{n}}[/tex]

[tex]\text{MOE}[/tex] = margin of error

[tex]\gamma[/tex] = confidence level

[tex]z_{\gamma}[/tex] = quantile

σ = standard deviation

n = sample size

As it is given that the sample size of the survey is 324, while the standard deviation of the survey is 1.6.

We know that the value of the z for 95% confidence interval is 1.96. Therefore, using the formula of the standard of error we can write it as,

[tex]\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\dfrac{\sigma^{2}}{n}}\\\\\text{MOE}_{\gamma}=1.96 \times \sqrt{\dfrac{1.6^{2}}{324}}\\\\\text{MOE}_{\gamma}=0.1742[/tex]

Hence, the margin of error for 324 adults surveyed with 1.6 standard deviations is 0.1742.

Learn more about Margin of Error:

https://brainly.com/question/6979326

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