reply to "Hello all, One for the mean price using a t interval. One for the proportion of cars priced below the mean using a z interval. I used the same ten vehicles as before and excluded the supercar. T interval for the mean. From Week 2 my sample mean was 41,760.40 dollars. In Week 4 I recalculated the sample standard deviation as 7,534.01 dollars. The sample size is n = 10, so degrees of freedom is 9. The standard error is 7,534.01 divided by the square root of 10 which is about 2,382.7. The t critical value for 95 percent with df = 9 is about 2.262. The margin of error is 2.262 times 2,382.7 which is about 5,388.6. The 95 percent t confidence interval for the mean price is about 36,372 dollars to 47,149 dollars. I used Excel functions AVERAGE, STDEV.S, and T.INV.2T to get these numbers. Z interval for the proportion below the mean. From Week 3 I defined success as a car priced below the mean. In my data 5 of 10 cars were below the mean, so p hat is 0.50 and q hat is 0.50. The z critical value for 95 percent is 1.96. The standard error is the square root of p hat times q hat over n which is the square root of 0.25 divided by 10 which is about 0.1581. The margin of error is 1.96 times 0.1581 which is about 0.3099. The 95 percent confidence interval for the true proportion of cars below the mean is about 0.19 to 0.81. What I think these intervals show. The t interval says a typical mean price for this set of vehicles likely falls somewhere in the mid to upper forty per