How to calculate Confidence Level and Sample Size?

Update: Corrected math error 6/5 = 1.2 and corrected final results.

Suppose X is a normal random variable:

standard deviation=0.05; n=9; empirical mean= 4.38

A) What is the confidence level of the interval [4.36, 4.40]?

Confidence Low = mean - Z_critical* Standard dev. / Sqrt(N)

Confidence High = mean + Z_critical*standard dev. /Sqrt(N)

Confidence High -mean = +Z_Critical*standard dev/ Sqrt(N)

4.40-4.38 = Z_Critical * 0.05/ sqrt(9) = Z_Critical* 0.05/3

0.02 = Z_critical*0.05/3

Z_Critical = 3*0.02/ 0.05 = 0.06/0.05 = 6/5 = 1.2

https://www.stat.tamu.edu/~lzhou/stat302/standardn...

so look P(z< 1.2) = .88493

so the range would be from z = -1.2 to z = 1.2

P(z< -1.1) (due to symmetry) = 1- .88493

so range is .88493 - (1 - .88493 )= 0.76986

then covert to percentage by multiplying by 100

=== answer (A)

100 * 0.76986= 76.986 Percent Confidence Range

Update: B) What should the sample size be if we wanted to know the mean at confidence level 99% with a margin of error

Margin of Error is half the confidence interval

Margin of Error = Z_critical* standard deviation/sqrt(N)

Margin of Error *sqrt(N) = Z_Critical*standard deviation

sqrt(N) = Z_critical*standard_deviation / Marigin of error

we know everything but Z_critical right off

sqrt(N) = Z_Critical *0.05/ 0.01 = 5*Z_critical

N = Z_Critical^2 * 5^2 = 25*Z_Critical

N= 25*Z_critical

Z_critical for 99 % confidence which goes from 0.5 to 0.995

would be the P(z< Z) = 0.995

find the two closest values and interpolate if necessary

P(z<2.57) = .99492

P(z<2.58) = .99506

P(z< Z) = 0.995

interpolating

2.57 + ( 0.995 - 0.99492) * 0.01 / (0.99506-0.99492)

2.57 + 0.00008 *0.01/ 0.00014 ) = 2.57 + 8/14*0.01 = 2.575714286

Z_Critical = approx . 2.575714286

N = 25* (2.575714286 )^2 = 165.857602

but N has to be integer

=== answer

N = 166

=== checking answer

Margin of Error = 2.575714286 * 0.05/sqrt(166) = 0.00999571

This is less than 0.01 so it is good.

.