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Sagot :
The sample size needed to provide a margin of error of 2 or less with a .95 probability when the population standard deviation equals 11 is 116.
Given margin of error of 2,probability of 0.95 and standard deviation is 11.
We have to calculate the sample size needed to provide a margin of error of 2 or less with a 0.95 probability when the population standard deviation of 11.
We know that the margin of error is the difference between the calculated values and actual values.
Margin of error=z*σ/[tex]\sqrt{n}[/tex]
where z is z value ,σ is standard deviation ,n is the sample size.
First we have to calculate the z value for the corresponding p value which is given of 0.95. We assume that the test is two tailed. z value is 1.96.
Now put the value of z , σ, n and margin of error in the formula of margin of error to get the value of n which is sample size.
2>=1.96*11/[tex]\sqrt{n}[/tex]
[tex]\sqrt{n}[/tex]>=(1.96*11)/2
[tex]\sqrt{n}[/tex]>=10.78
By squaring both sides we get,
n>=116.2064
Hence the sample size which is needed to provide a margin of error of 2 or less with a 0.95 probability when the population standard deviation equals 116.
Learn more about margin of error at https://brainly.com/question/10218601
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