The average life of a sample of 25 was 280 hours a Determine

The average life of a sample of 25  was 280 hours.

(a) Determine a 90% confidence interval estimate of the mean lifetime.

(b) How large a sample would be needed to obtain a 98% confidence interval of smaller size than the interval obtained in part (a)?

I understand (a), but especially I do not understand (b)..

Thank you so much.

Have a nice day!

Solution

a)
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=280
Standard deviation( sd )=60
Sample Size(n)=25
Confidence Interval = [ 280 ± Z a/2 ( 60/ Sqrt ( 25) ) ]
= [ 280 - 1.64 * (12) , 280 + 1.64 * (12) ]
= [ 260.32,299.68 ]

b)
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=280
Standard deviation( sd )=60
Sample Size(n)=25
Margin of Error = Z a/2 * 60/ Sqrt ( 25)
= 1.64 * (12)
= 19.68

Margin of error calculated in PART A is 19.68.

Minimum sample size to compute at 98% C.I is = n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.02% LOS is = 2.33 ( From Standard Normal Table )
Standard Deviation ( S.D) = 60
ME =19.68
n = ( 2.33*60/19.68) ^2
= (139.8/19.68 ) ^2
= 50.462 ~ 51      

At sample size 51, computed confidence interval is smaller size than the interval obtained in part (a)