Question

An urban planning group is interested in estimating the difference between mean household income for two neighborhoods in a large metropolitan area. Independent samples of house holds in the neighborhood provided the following results: Neighborhood Income (in thousands of dollars) 1 21.5 34.6 18.9 16.0 63.0 43.5 29.5 37.5 2 32.4 31.5 44.5 52.0 43.0 56.4 42.5 34.7 42.2 44.6 a. What function would be used to construct a confidence interval for the difference in Neighborhood 1 and Neighborhood 2's mean income? b. Construct a 97.5% confidence interval for the difference in Neighborhood 1 and Neighborhood 2's mean income. Round your answers to two decimal places. ( C. On average, which neighborhood has a higher income?

Answer 1

Solution:

#R version 3.3.2 n1=c(21.5,34.6,18.9,16,63,43.5,29.5,37.5) length(n1) mean(n1) sd(n1) n2=c(32.4,31.5,44.5,52,43,56.4,42.5,34.7, 42.2,44.6) length(n2) mean(n2) sd(n2) Save it Put on a wall Run it (Shift+Enter) [ + ] Show input Absolute running time: 0.47 sec, cpu time: 0.9 sec, memory peak: 36 Mb, absolute service time: 0,66 sec [1] 8 [1] 33.0625 [1] 15.42419 [1] 10 [1] 42.38 [1] 7.999417

Si = 15.4242 Given: ni =8 X = 33.0625 X2 =42.38 7. 9994 S2 = 12 =10 interval is, 97.5% confidence s2 (XI-X2 + +*2 + Du h2 (33.0625 - 842.38) 1 8,4729 (15.4242) 2 (7.9994) f 8 St* = 2-4729, at 4/2 = 0.0125 df = n 182-2 = 16 from t-tabic - 9. 3175 † 14.8657 5.55) The 99.7.5% confidence interval is (-24.18 co) has higher income. average, neighborhood 2