Question

MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER 2. [-12 Points) DETAILS DIEZSTAT3 5.E.030. Consider the diamond prices (standardized by weight) for diamonds with weights 0.99 carat and 1 carat. See the table for summary statistics, and then construct a 95% confidence interva for the average difference between the standardized prices of 0.99 and 1 carat diamonds. You may assume the conditions for inference are met. (Round your answers to two decimal places.) 0.99 carat 1 carat Mean $43.53 $56.81 SD $13.32 $15.11 23 23 n $ You may need to use the appropriate table in the Appendix of Tables to answer this question T Additional Materials Book

Answer 1

so! let us denote by I the 0.99 Carat I caror and by a the Then diamonds. we are given that N, = 43.53 S, - 13.32 and n, =23 le la = 56.81 S, - 15.10 and n₂ =23 Since we are assuming the population Variances equal we need to get the estimat variance of the pooled as follows o (1,-1). $;" +(12-1) 5, s U tn 2 -2 (23-1) (13.32) + (23-1) (15.11) 11 23+23–2 az s T 202.86.73 S=14.243.) thena the standard error estimate for difference in mean is (ส - 14.2431*5 1/4 + 1 syttat se: s Se=4.2003 CS$canned with CamScanner

we know interval that the 95%. Confidina for the difference in population the formula means is given by ( ā-5.) I terale X se in, to₂-2 where tinda I mit ne-a is the critical value of t. For aur problem = 2.0514 0.975; tog Substituting the valus we get CI - (43.53-56.81) † 2.0514 X 4.2003. -13.28 £ 8.6165 Thi = (-21.8965, -4.6635) 95%. confidence Interned for the difference between the standardino Carat daimonda average Prices of 0.99 and I (-21.8965, -4.6635) CSscanned with CamScanner