# A sample of 30 observation has

A sample of 30 observation has sample mean x̅ = 53 and samplestandard deviation of s = 3. Construct a 99% two sided confidenceinterval for the population mean. If we want the length of 99% twosided confidence interval to be smaller than 1, how many moresamples do we need to take?

Please show your work, I will be studying the solution step bystep.

Solution:

Given:

Sample size = n = 30

Sample mean =

Sample Standard Deviation = s = 3

Confidence level = c = 99% = 0.99

Part a) Find 99% two sided confidence intervalfor the population mean.

where

where tc is t critical value for c = 99% confidencelevel.

df = n – 1 = 30 – 1 = 29

two tail area = 1 – c = 1 – 0.99 = 0.01

From t table , we get:

tc = 2.756

Thus

Thus

Thus 99% two sided confidence interval for the population meanis between the limits: (51.49 , 54.51 )

Part b) we want the length of 99% two sidedconfidence interval to be smaller than 1

That is length = 1

then E = Margin of Error = length / 2 = 1 / 2 = 0.5

Formula for sample size n is:

Zc is z critical value for c = 0.99 confidencelevel.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area andfind corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and0.9951 and both are at same distance from 0.9950, Hencecorresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Zc = 2.575

Thus

We have to find: how many more samples do we need to take?

Thus 239 – 30 = 209

Thus we need to take 209 more samples

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