Question

A real-estate company appraised the market value of 36 homes in Lyttelton and found that the sample mean and standard deviation were $150,000 and $17,000 respectively. The real-estate company also appraised the market value of 45 homes in Aranui and found that the sample mean and standard deviation were $100,000 and $12,000 respectively. Calculate the 90% confidence interval estimate for the population difference in market value between the Lyttelton and Aranui areas .

Answer 1

Sample size Sample mean Sample S.D.

Sample 1
`$n_1=36$`
`$\bar{x}_1=150,000$`
`$s_1=17,000$`

Sample 2
`$n_2=45$`
`$\bar{x}_2=100,000$`
`$s_2=12,000$`

`$df = (\left((s_1^2)/(n_1)+(s_2^2)/(n_2)\right)^2)/((1)/(n_1-1)\left((s_1^2)/(n_1)\right)^2+(1)/(n_2-1)\left((s_2^2)/(n_2)\right)^2)$`

= 60

Therefore, significance level, α = 0.1

Critical value, t* = 1.6706

So, the margin of error,
`$t^*=\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}$`

= 559.9896

Lower limit,
`$(\bar x_1-\bar x_2)-\text{(margin of error)}=44402.0104$`

Upper limit,
`$(\bar x_1-\bar x_2)+\text{(margin of error)}=55597.9896$`

Therefore 90% C.I. is (44402.0104, 55597.9896) or
`$44402.0104 < \mu_1 - \mu_2 < 55597.9896$`