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ELEN [110]
4 years ago
12

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro

m a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.2 hours. Calculate the margin of error (to two decimals) for a confidence interval with a confidence coefficient of 0.95.
Mathematics
1 answer:
VladimirAG [237]4 years ago
4 0

Answer:

Margin of error  for a 95% of confidence intervals is 0.261

Step-by-step explanation:

<u>Step1:-</u>

 Sample n = 81 business students over a one-week period.

 Given the population standard deviation is 1.2 hours

 Confidence level of significance = 0.95

 Zₐ = 1.96

Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }

Given n=81 , σ =1.2 and  Zₐ = 1.96

<u>Step2:-</u>

<u />Margin of error (M.E) = \frac{Z_{\alpha  }S.D }{\sqrt{n} }<u />

<u />Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }<u />

On calculating , we get

Margin of error = 0.261

<u>Conclusion:-</u>

Margin of error  for a 95% of confidence intervals is 0.261

<u />

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