Question

To find out / estimate the average value of the nominal shear strength of the steel beams in kN of all production, a test of 15 steel beam samples is carried out. The result is as follows:

580, 400, 428, 825, 850,875, 920, 550, 575, 750, 636, 360, 590, 735, 950

Estimate the average price (interval) of the steel beam population with a 99% confidence interval.

Answer 1

Given: n= 15

we can find mean and sd from given data.

Using ti84, sample mean ,$\overline{x} = 668.27$ , sample sd s=192.09

Formula:

Since population sd is unknown, we use t- confidence interval for population mean.

$[\overline{x}-t^{*}\frac{s}{\sqrt{n}},\overline{x}+t^{*}\frac{s}{\sqrt{n}}]$

The number of degrees of freedom are df = n-1 =15 - 1 = 14

and the significance level is alpha = 0.99Based on the provided information, the critical t-value for α=0.99 and df = 14 is t*= 0.013

$[668.27-0.013*\frac{192.09}{\sqrt{15}},668.27+0.013*\frac{192.09}{\sqrt{15}}]$

simplifying , required confidence interval is =(667.633,668.899)