Assume the measurements of nominal shear strength (in kN) are normally distributed. To estimate the mean...

Assume the measurements of nominal shear strength (in kN) are normally distributed. To estimate the mean of this distribution, a sample of 26 measurements was collected with sample mean 678.27 and the sample standard deviation 82.89. Find a 90% confidence interval. Please show work. Thanks.

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Expert Solution

Here, Population standard deviation is not given, then we will use student t distribution
We want to find 90% confidence interval for population mean μ
The formula of confidence interval for population mean μ is as follow:

Lower limit =
Upper limit =
Where E is the margin of error
the formula of margin of error is as follows:

given information is -

For confidence level =0.1

(I have used excel to obtain this, command is "=tinv(probability,df), here, probability =0.1 , df =25)

Therefore,

Lower limit = = 678.27 - 27.77 =650.50
Upper limit = =678.27 + 27.77 =706.04
Thus, 90% confidence interval for population mean μ is (650.50 ,706.04 )

Interpretation: We are 90% confident that this interval will include population mean μ.

(Please give thumps up, if you like the answer )

orchestra answered 1 year ago

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