m e a n = 1 + 5 + 2 + 2 + 4 + 1 + 2 + 2 + 4 + 3 10 = 2.6 mean=\dfrac{1+5+2+2+4+1+2+2+4+3}{10}=2.6 m e an = 10 1 + 5 + 2 + 2 + 4 + 1 + 2 + 2 + 4 + 3 = 2.6 s 2 = ∑ ( x i − x ˉ ) 2 n − 1 = 16.4 9 s^2=\dfrac{\sum{(x_i-\bar{x})^2}}{n-1}=\dfrac{16.4}{9} s 2 = n − 1 ∑ ( x i − x ˉ ) 2 = 9 16.4
s = s 2 = 2 4.1 3 ≈ 1.3499 s=\sqrt{s^2}=\dfrac{2\sqrt{4.1}}{3}\approx1.3499 s = s 2 = 3 2 4.1 ≈ 1.3499 a. The number of degrees of freedom are d f = 10 − 1 = 9 , df=10-1=9, df = 10 − 1 = 9 , and the significance level is α = 0.01. \alpha=0.01. α = 0.01.
The critical t-value for α = 0.01 \alpha=0.01 α = 0.01 and df = 9 degrees of freedom is t c = 3.249823. t_c=3.249823. t c = 3.249823.
The 99% confidence for the population mean μ \mu μ is computed using the following expression
C I = ( x ˉ − t c × s n , x ˉ + t c × s n ) = CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})= C I = ( x ˉ − t c × n s , x ˉ + t c × n s ) =
= ( 2.6 − 3.2498 × 1.3499 10 , 2.6 + 3.2498 × 1.3499 10 ) = =(2.6-3.2498\times \dfrac{1.3499}{\sqrt{10}}, 2.6+3.2498\times \dfrac{1.3499}{\sqrt{10}})= = ( 2.6 − 3.2498 × 10 1.3499 , 2.6 + 3.2498 × 10 1.3499 ) =
= ( 1.2127 , 3.9873 ) =(1.2127, 3.9873) = ( 1.2127 , 3.9873 )
b. The number of degrees of freedom are d f = 10 − 1 = 9 , df=10-1=9, df = 10 − 1 = 9 , and the significance level is α = 0.05. \alpha=0.05. α = 0.05.
The critical t-value for α = 0.05 \alpha=0.05 α = 0.05 and df = 9 degrees of freedom is t c = 2.262156. t_c=2.262156. t c = 2.262156.
The 95% confidence for the population mean μ \mu μ is computed using the following expression
C I = ( x ˉ − t c × s n , x ˉ + t c × s n ) = CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})= C I = ( x ˉ − t c × n s , x ˉ + t c × n s ) =
= ( 2.6 − 2.262156 × 1.3499 10 , 2.6 + 2.262156 × 1.3499 10 ) = =(2.6-2.262156\times \dfrac{1.3499}{\sqrt{10}}, 2.6+2.262156\times \dfrac{1.3499}{\sqrt{10}})= = ( 2.6 − 2.262156 × 10 1.3499 , 2.6 + 2.262156 × 10 1.3499 ) =
= ( 1.6343 , 3.5657 ) =(1.6343, 3.5657) = ( 1.6343 , 3.5657 )
Comments
Leave a comment