how can you respond to this discussion: I computed sample means for four disjoint subsamples from my Part 1 random sample (n = 10, 30, 50, 100) and used the known population standard deviation = $704,500 to build 95% confidence intervals. For a 95% CI we use z = 1.96 and the formula CI = x 1.96 ( n ) . CI= x 1.96( n ). Below is the table template I used (fill in your sample means). The standard errors and 95% margins (calculated with = $704,500) are provided so you can paste your x x and get the interval immediately. n sample mean x x standard error / n / n 95% margin (1.96SE) 95% CI 10 $___________ $222,782.46 $436,653.62 x 436 , 653.62 x 436,653.62 30 $___________ $128,623.51 $252,102.09 x 252 , 102.09 x 252,102.09 50 $___________ $99,631.35 $195,277.44 x 195 , 277.44 x 195,277.44 100 $___________ $70,450.00 $138,082.00 x 138 , 082.00 x 138,082.00 Example interpretation (n = 100): if x = $ 480 , 000 x =$480,000, the 95% CI = $480,000 $138,082 ($341,918, $618,082). This means I am 95% confident the true mean home price in the population lies between $341,918 and $618,082 (given normality and known ). Larger samples tend to produce sample means that vary less and give narrower confidence intervals because SE = /n shrinks as n grows. Reporting sample size alongside statistics helps audiences judge reliability; a mean from n=10 should be treated more cautiously than one from n=100. and Compare your 95% confidence interval for n = 100 with that of your peers. Additionally, comment on the role that sample