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

Solution :

                    Sample size       Sample mean             Sample S.D.

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

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

$df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}$

   = 60

Therefore, significance level, α = 0.1

Critical value, t* = 1.6706

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

                                          = 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 $44402.0104 < \mu_1 - \mu_2 < 55597.9896$