In a sample of 100 steel wires the average breaking strength is 45 kN, with a st

Question

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?

Explanation / Answer

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 ]

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