8:The mean and standard deviation of the maximum loads supported by 60 cables are 11.09 tons and 0.73 tons, respectively. Find 95% confidence limits for the mean
Answers
Step-by-step explanation:
The 95% confidence interval is 10.91 < 11.28
The 99% confidence interval is 10.85 < 11.33
Step-by-step explanation:
From the question we are told that
The sample mean is \= x= 11.09\
cons
The standard deviation is o =
0.73 tons
The sample size is n =60
From the question we are told the confidence level is 95%, hence the evel of significance is
α= (100 – 95)%
<= a = = 0.05
Generally from the normal distribution able the critical value of is
1.96
Generally the margin of error is nathematically represented as
E = Z₂ * J σ
<= E = 1.96* \frac{ 0.73 }{\sqrt{n60}
=>
E = 0.1847
Generally 95% confidence interval is mathematically represented as
\= x-E < \mu < \=x +E
=> 11.09 0.1847 << 11.09 +
0.1847
=>
10.91 << 11.28
From the question we are told the confidence level is 99%, hence the evel of significance is
α = (100 - 99)%
= 0.01/tex] </p><p> Generally fromthenormaldistributiont S
a =
= 2.58
Generally the margin of error is mathematically represented as
E = Z *
<= E = 2.58* \frac{ 0.73 }{\sqrt{n60}
=> E = 0.2431
Generally 95% confidence interval is mathematically represented as
\= x -E < \mu < \=x +E
=> 11.09 0.2431 -0.2431 < µ< 11.09 +
<=
10.85< < 11.33