In a sample of 100 steel wires the average breaking strength

In a sample of 100 steel wires the average breaking strength is 45 kN, with a standard deviation of 4 kN.

A. If you know the population distribution, what is the distribution of the sample statistic?

B. How many wires must be sampled so that a 95% confidence interval specifies the mean breaking strength to within ±0.3kN? Use s as an estimate for .

C. Find the 75%,85% and 95% confidence interval for n = 100 for the mean breaking strength of this type of wire. What do you notice about the width of the CI as the confidence level increases?

Solution

B)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 4
ME =0.3
n = ( 1.96*4/0.3) ^2
= (7.84/0.3 ) ^2
= 682.951 ~ 683      

                  
C)AT 0.75
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=45
Standard deviation( sd )=4
Sample Size(n)=100
Confidence Interval = [ 45 ± t a/2 ( 4/ Sqrt ( 100) ) ]
= [ 45 - 1.157 * (0.4) , 45 + 1.157 * (0.4) ]
= [ 44.537,45.463 ]
                  
AT 0.85
Confidence Interval = [ 45 ± t a/2 ( 4/ Sqrt ( 100) ) ]
= [ 45 - 1.451 * (0.4) , 45 + 1.451 * (0.4) ]
= [ 44.42,45.58 ]

AT 0.95
Confidence Interval = [ 45 ± t a/2 ( 4/ Sqrt ( 100) ) ]
= [ 45 - 1.984 * (0.4) , 45 + 1.984 * (0.4) ]
= [ 44.206,45.794 ]