Question

In Exercise 5.28, we discussed diamond prices (standardized by weight) for diamonds with weights 0.99 carats and 1 carat. See the table for summary statistics, and then construct a 95% confidence interval for the average difference between the standardized prices of 0.99 and 1 carat diamonds. You may assume the conditions for inference are met.

Answer 1

Let \(x 1\) denotes 1 carats and \(\times 2\) denotes \(0.99\) carats of diamond differnce in sample mean is

$$ \begin{aligned} &\bar{x} 1-\bar{x} 2 \\ &=56.81-44.51 \\ &=12.3 \end{aligned} $$

pooled sample standard deviation is

$$ \begin{aligned} &s p=\sqrt{\left[(n 1-1) s 1^{2}+(n 2-1) s 2^{2}\right] /(n 1+n 2-2)} \\ &=\sqrt{\left[(22) * 16.13^{2}+(22) * 13.32^{2}\right] /(23+23-2)} \\ &=\sqrt{9627.18 / 44} \\ &=14.79 \end{aligned} $$

standard error is

$$ \begin{aligned} &\text { S.E. }=s p * \sqrt{(1 / n 1)+(1 / n 2)} \\ &=14.79 * \sqrt{(1 / 23)+(1 / 23)} \\ &=4.36 \end{aligned} $$

at \(95 \%\) confidence interval \(Z(\) alph \(a / 2)=1.96\)

confidence interval is

$$ \begin{aligned} &(\bar{x} 1-\bar{x} 2) \pm Z(\alpha / 2) * S . E \\ &=\left(12.3-1.96^{*} 4.36,12.3+1.96 * 4.36\right) \\ &=(12.3-8.55,12.3+8.55) \\ &=(3.75,20.85) \end{aligned} $$