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MissTica
3 years ago
8

There were 324 adults surveyed. Among the participants, the mean number of hours of sleep each night was 7. 5 and the standard d

eviation was 1. 6. The margin of error, assuming a 95% confidence level, is approximately. Round to the nearest hundredth.
Mathematics
1 answer:
NeTakaya3 years ago
5 0

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

<h3>What is the margin of error?</h3>

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,

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

\text{MOE} = margin of error

\gamma = confidence level

z_{\gamma} = 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,

\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

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

Learn more about Margin of Error:

brainly.com/question/6979326

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Length of string used by Claire = 7 / 3 feet

Step-by-step explanation:

Given:

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Length of string Claire in starts = 4\frac{1}{6} feet = 25 / 6 feet

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Length of string used by Claire = Length of string Claire in starts - Length of string Claire in end

Length of string used by Claire = [25 / 6] - [11 / 6]

BY taking LCM

Length of string used by Claire = [25 - 11] / 6

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3 years ago
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