Question

Hexavalent chromium has been identified as an inhalation carcinogen and an air toxin of concern in a number of different locales. An article gave the accompanying data on both indoor and outdoor concentration (nanograms/m³) for a sample of houses selected from a certain region.

	House
	1	2	3	4	5	6	7	8	9
Indoor	0.11	0.12	0.13	0.16	0.16	0.16	0.17	0.18	0.19
Outdoor	0.25	0.64	0.43	0.50	0.93	0.31	0.45	0.80	0.82

	House
	10	11	12	13	14	15	16	17
Indoor	0.19	0.21	0.21	0.22	0.22	0.22	0.22	0.23
Outdoor	0.24	0.28	0.28	1.51	0.62	0.25	0.17	0.98

	House
	18	19	20	21	22	23	24	25
Indoor	0.24	0.26	0.26	0.27	0.27	0.29	0.30	0.32
Outdoor	1.55	0.86	0.48	0.08	0.50	0.84	0.45	1.20

	House
	26	27	28	29	30	31	32	33
Indoor	0.32	0.33	0.38	0.43	0.44	0.49	0.58	0.66
Outdoor	0.44	0.23	0.33	1.22	0.66	0.72	0.95	0.32

Calculate a confidence interval for the population mean difference between indoor and outdoor concentrations, μ_indoor − μ_outdoor, using a confidence level of 95%. (Round your answers to two decimal places.)

Interpret the interval.

(b) If a 34th house were to be randomly selected from the population, between what values would you predict the difference in concentrations to lie? (Round your answers to two decimal places.)

Answer 1

Indoor	Outdoor	Difference
0.11	0.25	-0.14
0.12	0.64	-0.52
0.13	0.43	-0.30
0.16	0.50	-0.34
0.16	0.93	-0.77
0.16	0.31	-0.15
0.17	0.45	-0.28
0.18	0.80	-0.62
0.19	0.82	-0.63
0.19	0.24	-0.05
0.21	0.28	-0.07
0.21	0.28	-0.07
0.22	1.51	-1.29
0.22	0.62	-0.40
0.22	0.25	-0.03
0.22	0.17	0.05
0.23	0.98	-0.75
0.24	1.55	-1.31
0.26	0.86	-0.60
0.26	0.48	-0.22
0.27	0.08	0.19
0.27	0.50	-0.23
0.29	0.84	-0.55
0.30	0.45	-0.15
0.32	1.20	-0.88
0.32	0.44	-0.12
0.33	0.23	0.10
0.38	0.33	0.05
0.43	1.22	-0.79
0.44	0.66	-0.22
0.49	0.72	-0.23
0.58	0.95	-0.37
0.66	0.32	0.34

Sample mean of the difference using excel function AVERAGE(), x̅_d = -0.3439

Sample standard deviation of the difference using excel function STDEV.S(), s_d = 0.3868

Sample size, n = 33

a) 95% Confidence interval :

At α = 0.05 and df = n-1 = 32, two tailed critical value, t-crit = T.INV.2T(0.05, 32) = 2.037

Lower Bound = x̅_d - t-crit*s_d/√n = -0.3439 - 2.037 * 0.3868/√33 = -0.48

Upper Bound = x̅_d + t-crit*s_d/√n = -0.3439 + 2.037 * 0.3868/√33 = -0.21

-0.48 < µ_d < -0.21

b) 95% prediction interval :

At α = 0.05 and df = n-1 = 32, two tailed critical value, t-crit = T.INV.2T(0.05, 32) = 2.037

Lower Bound = x̅_d - t-crit*s_d/√(1+1/n) = -0.3439 - 2.037 * 0.3868/√(1+1/33) = -1.12

Upper Bound = x̅_d + t-crit*s_d/√(1+1/n) = -0.3439 + 2.037 * 0.3868/√(1+1/33) = 0.43

-1.12 < µ_d < 0.43