Answer to Question #141661 in Statistics and Probability for Bell Locsin

Question #141661
A veterinarian in the barangay wanted to determine the average number of pets. They got the following values.
1 5 2 2 4 1 2 2 4 3
a. Determine the 99% confidence interval for the population mean.
b. Determine the 95% confidence interval for the population mean
1
Expert's answer
2020-11-02T20:04:28-0500
mean=1+5+2+2+4+1+2+2+4+310=2.6mean=\dfrac{1+5+2+2+4+1+2+2+4+3}{10}=2.6s2=(xixˉ)2n1=16.49s^2=\dfrac{\sum{(x_i-\bar{x})^2}}{n-1}=\dfrac{16.4}{9}

s=s2=24.131.3499s=\sqrt{s^2}=\dfrac{2\sqrt{4.1}}{3}\approx1.3499

a. The number of degrees of freedom are df=101=9,df=10-1=9, and the significance level is α=0.01.\alpha=0.01.

The critical t-value for α=0.01\alpha=0.01 and df = 9 degrees of freedom is tc=3.249823.t_c=3.249823.

The 99% confidence for the population mean μ\mu  is computed using the following expression


CI=(xˉtc×sn,xˉ+tc×sn)=CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})=

=(2.63.2498×1.349910,2.6+3.2498×1.349910)==(2.6-3.2498\times \dfrac{1.3499}{\sqrt{10}}, 2.6+3.2498\times \dfrac{1.3499}{\sqrt{10}})=

=(1.2127,3.9873)=(1.2127, 3.9873)

b. The number of degrees of freedom are df=101=9,df=10-1=9, and the significance level is α=0.05.\alpha=0.05.

The critical t-value for α=0.05\alpha=0.05 and df = 9 degrees of freedom is tc=2.262156.t_c=2.262156.

The 95% confidence for the population mean μ\mu  is computed using the following expression


CI=(xˉtc×sn,xˉ+tc×sn)=CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})=

=(2.62.262156×1.349910,2.6+2.262156×1.349910)==(2.6-2.262156\times \dfrac{1.3499}{\sqrt{10}}, 2.6+2.262156\times \dfrac{1.3499}{\sqrt{10}})=

=(1.6343,3.5657)=(1.6343, 3.5657)


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