1. How large a sample is required if we want to have 95% confidence that the ave

Question

1. How large a sample is required if we want to have 95% confidence that the average hardness of a sample of shafts is within 0.3 (margin of error) of the expected value if the variance is known to be 1.0, 2.0, 3.0, 4.0, 5.0, 6.0? Using the Excel spreadsheet to plot the relationship between the sample size (Y-axis) vs. standard deviation ().

2. If the observed concentration readings are 1.27, 1.33, 1.19, and 1.24, determine 95% confidence interval for the mean concentration assuming variance = 0.003.

3. Prove: E(S^2) = ^2, please show each step to demonstrate how you prove it.

Explanation / Answer

Q1.
When variance is 1.0
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) = 1
ME =0.3
n = ( 1.96*1/0.3) ^2
= (1.96/0.3 ) ^2
= 42.68 ~ 43  
When variance is 2.0  
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) = 1.4142135623731
ME =0.3
n = ( 1.96*1.4142135623731/0.3) ^2
= (2.77/0.3 ) ^2
= 85.37 ~ 86  
When variance is 3.0
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) = 1.73205080756888
ME =0.3
n = ( 1.96*1.73205080756888/0.3) ^2
= (3.39/0.3 ) ^2
= 128.05 ~ 129      
When variance is 4.0
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) = 2
ME =0.3
n = ( 1.96*2/0.3) ^2
= (3.92/0.3 ) ^2
= 170.74 ~ 171      
When variance is 5.0
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) = 2.23606797749979
ME =0.3
n = ( 1.96*2.23606797749979/0.3) ^2
= (4.38/0.3 ) ^2
= 213.42 ~ 214      
When variance is 6.0
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) = 2.44948974278318
ME =0.3
n = ( 1.96*2.44948974278318/0.3) ^2
= (4.8/0.3 ) ^2
= 256.11 ~ 257      
Q2.
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)=1.2575
Standard deviation( sd )=0.05477
Sample Size(n)=4
Confidence Interval = [ 1.2575 ± Z a/2 ( 0.05477/ Sqrt ( 4) ) ]
= [ 1.2575 - 1.96 * (0.03) , 1.2575 + 1.96 * (0.03) ]
= [ 1.2,1.31 ]

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