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How large a sample is needed in Question 1 if we wish to be 98% confident that our sample mean will be within 12 hours of the true mean? Use the standard deviation provided in Question 1.

User Marsolmos
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4 votes

Answer:

n=61

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We assume the following info:


\bar X=780 represent the sample mean


\sigma=40 represent the population standard deviation

n represent the sample selected (variable of interest)


\alpha=0.02 significance level

Confidence =0.98 or 98%

Me= 12 represent the margin of error required


ME=z_(\alpha/2)(\sigma)/(√(n)) (a)

And on this case we have that ME =12 and we are interested in order to find the value of n, if we solve n from equation (a) we got:


n=((z_(\alpha/2) \sigma)/(ME))^2 (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got
z_(\alpha/2)=2.33, replacing into formula (b) we got:


n=((2.33(40))/(12))^2 =60.32 \approx 61

So the answer for this case would be n=61 rounded up to the nearest integer

User Rick Love
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