if we are very certain that the true standard deviation of the number of defects per screen is below 2, what sample size would be required so that the width of the 95% confidence interval for the mean number of defects per screen is at most 0.34? make sure you enter a whole number below. 532 correct: your answer is correct.

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27-10-2022
Mathematics

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if we are very certain that the true standard deviation of the number of defects per screen is below 2, what sample size would be required so that the width of the 95% confidence interval for the mean number of defects per screen is at most 0.34? make sure you enter a whole number below. 532 correct: your answer is correct.

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qwterabite qwterabite · Answer 1 · 2022-11-01T10:32:15-04:00

The sample size would be 133 for the width of the 95% confidence interval for the mean number of defects per screen is at most 0.34

The margin of error also called the confidence interval, tells you how much you'll anticipate your overview comes about to reflect the sees of the overall population. Keep in mind that surveying could be an adjusting act where you utilize a littler group to speak to a much bigger one

The formula for calculating margin error is

E= Z*(standard deviation /set(sample size ))

where Z is the z score.

Since we are provided with a standard deviation which is 2 and a confidence interval 95% and a mean defect which is 0.34 and for the confidence 95% the z score is 1.96.

So, the sample size = (Z. Sd /E)^2

= (1.96*2/0.34)^2

= 132.92

= 133

the sample size will be 133

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