Answer:
a) The Margin of error = 2.6723
b)<em> The Standard error = 1.1299</em>
Step-by-step explanation:
Given data
x 109 105 112 110 115 106 108 109
Mean of x = ![\frac{109 +105 + 112 + 110 +115+ 106 +108+ 109}{8} = 109.25](https://tex.z-dn.net/?f=%5Cfrac%7B109%20%2B105%20%2B%20112%20%2B%20110%20%2B115%2B%20106%20%2B108%2B%20109%7D%7B8%7D%20%3D%20109.25)
x⁻ = 109.25
x : 109 105 112 110 115 106 108 109
x- x⁻ : -0.25 -4.25 2.75 0.75 5.75 -3.25 -1.25 -0.25
(x-x⁻)²: 0.062 18.062 7.562 0.562 33.062 10.562 1.562 0.062
∑((x-x⁻)² = 71.5
![S^2 = \frac{71.5}{8-1} = \frac{71.5}{7} =10.2142](https://tex.z-dn.net/?f=S%5E2%20%3D%20%5Cfrac%7B71.5%7D%7B8-1%7D%20%3D%20%5Cfrac%7B71.5%7D%7B7%7D%20%3D10.2142)
The Standard deviation of sample S = 3.1959
a)
The Margin of error is determined by
![M.E = t_{\frac{\alpha }{2} } \frac{ S}{\sqrt{n} }](https://tex.z-dn.net/?f=M.E%20%3D%20t_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%20%5Cfrac%7B%20S%7D%7B%5Csqrt%7Bn%7D%20%7D)
Degrees of freedom γ =n-1 = 8-1 =7
![t_{\frac{\alpha }{2} } = t_{\frac{0.05}{2} } = t_{0.025} = 2.365](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%20%3D%20t_%7B%5Cfrac%7B0.05%7D%7B2%7D%20%7D%20%3D%20t_%7B0.025%7D%20%3D%20%202.365)
![M.E = 2.365{ \frac{ 3.195}{\sqrt{8} }](https://tex.z-dn.net/?f=M.E%20%3D%20%202.365%7B%20%5Cfrac%7B%203.195%7D%7B%5Csqrt%7B8%7D%20%7D)
<em>The Margin of error = 2.6723</em>
<em>b) The Standard error is determined by</em>
![S.E = \frac{ S}{\sqrt{n} }](https://tex.z-dn.net/?f=S.E%20%3D%20%5Cfrac%7B%20S%7D%7B%5Csqrt%7Bn%7D%20%7D)
![S.E = \frac{ S}{\sqrt{n} } = \frac{3.1959}{\sqrt{8} } = 1.1299](https://tex.z-dn.net/?f=S.E%20%3D%20%5Cfrac%7B%20S%7D%7B%5Csqrt%7Bn%7D%20%7D%20%3D%20%5Cfrac%7B3.1959%7D%7B%5Csqrt%7B8%7D%20%7D%20%3D%201.1299)
<em>The Standard error = 1.1299</em>