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There are 2,367 teachers in a certain school district. 600 were polled asking if they felt computers in the classroom were beneficial to. 85% of the teachers polled felt that computers in the classroom were beneficial to students. What is the confidence interval for the population mean of teachers in this district that feel computers in the classroom are beneficial to students, using a 99% confidence level?

(81.25%, 88.75%)

(81.98%, 88.21%)

(82.14%, 87.86%)

(82.60%, 87.40%)

A confidence interval is how much instability there's with any specific statistic. Confidence intervals are frequently utilized with a margin of error. It tells you how confident you'll be able be that the results from a poll or survey reflect what you'd expect to find in the event that it were possible to survey the whole population. Confidence intervals are inherently associated to confidence levels.

Here the sample size, n=600

85% of the teachers polled felt that computers in the classroom were beneficial to students.

so the proportion=85%=0.85

Confidence level=99%

so. the level of significance, α=1-0.99=0.01

So, z critical value at 99% confidence level,

[tex]Z_{\frac{\alpha }{2} } =2.576[/tex]

So the formula to find confidence interval is,

[tex]proportion-Z_\frac{\alpha }{2} *\sqrt{\frac{prop.(1-prop)}{n} } < p < proportion+Z_\frac{\alpha }{2} *\sqrt{\frac{prop.(1-prop)}{n} }[/tex]

[tex]0.85-2.576*\sqrt{\frac{0.85*(1-0.85)}{600} } < p < 0.85+2.576*\sqrt{\frac{0.85*(1-0.85)}{600} }[/tex]

=(0.8125,0.8875)

=(81.25%,88.75%)

To learn more about the confidence interval please follow: https://brainly.ph/question/21549681

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