Math, asked by mishika6692, 1 day ago

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

Answered by manjuagrawal19nov
0

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

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