Question

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

Answer 1

$$mean=\dfrac{1+5+2+2+4+1+2+2+4+3}{10}=2.6$$ $$s^2=\dfrac{\sum{(x_i-\bar{x})^2}}{n-1}=\dfrac{16.4}{9}$$

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

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

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

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

$$CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})=$$

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

$$=(1.2127, 3.9873)$$

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

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

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

$$CI=(\bar{x}-t_c\times \dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times \dfrac{s}{\sqrt{n}})=$$

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

$$=(1.6343, 3.5657)$$