If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.

Given 95% confidence level, standard deviation=0.05.

We know that margin of error is the range of values below and above the sample statistic in a confidence interval.

We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.

We know that margin of error for a confidence interval is given by:

Me= $z/2* ST/\sqrt{N}$

α=1-0.95=0.05

α/2=0.025

z with α/2=1.96 (using normal distribution table)

Solving for n using formula of margin of error.

$n=(z/2ST/Me)^{2}$

n= $(1.96*0.05)^{2} /(0.01)^{2}$

=96.4

By rounding off we will get 97.

Hence the sample size required will be 97.

Learn more about standard deviation at brainly.com/question/475676

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The given question is incomplete and the full question is as under:

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?

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2 years ago