Question

A random sample of n, - 11 winter days in Denver gave a sample mean pollution Index x - 43. Previous studies show that 0,- 14. For Englewood (a suburb of Denver), a random sample of n, - 17 winter days gave a sample mean pollution Index of xy - 51. Previous studies show that 0, - 16. Assume the pollution Index is normally distributed in both Englewood and Denver, Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a 1% level of significance. (a) What is the level of significance? 0.01 State the null and alternate hypotheses. O Mo:My My My! My > M2 O Ho: M1 - HH, + H2 O Ho: My <H2; H, H, My O Mo: , MH, H, H2 (b) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with known standard deviations The Student's t. We assume that both population distributions are approximately normal with known standard deviations The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations What is the value of the sample test statistic (Test the difference - Round your answer to two decimal places.) (c) Find (or estimate the value. (Round your answer to four decimal places.) 03355 X Sketch the sampling distribution and show the area corresponding to the value

Answer 1

Solution n = 11, D₂ = 17 n = 43 = 51 6 = 14 52 = 16 (a) level of significance = 0.01 (-) Ho! 4 = 42 ; Hiritu2 (6) tax-X2 -14 - H2) u N10.1) since, 5, 62 are known, [we could use to if we don't know the 61, 62 and both me population & normally as follows normal dinsitution.] value of the sample test statistic 16. Mr=42 a value of sample Test Stalish! 143-511 = 1.39 View the (C) p-values - P(1tly 1:39) = P(+) 139) + P(+6139) a zp(+7139) - 2 [PLETO) - P/OLE (139)] = 2(0.5 0'4177] = 0'16461 unde Ho ă, căz-14, 2) 114 129 0 139 The conca of the shaded region in Pivale

n = 41 12 = 40 2 - 49 = 4:0 612 30 62-13 (A) level of singrisian = 0.05 (#) Hon=42 vs HIM,+"2 u. 22 - (-2) (011) (6) ta - 12 since, 6.52 are kurson and on boter population is normal. [since 61,62 ane knnon we we chave tuNloi) but if 66 are unknoon thom it would be tnitn₂-2 . Ho! Mg =42) Value & (under Test atahetic - n- 2.- (-42) & 2406] 4.9-4.0 – 2.406. (c) P-value & P (2) 72.406) = P((A7 2:41) - Plt) 2:41)+P(+(-2:41) - 2 P(+) 2:41) = 2[PC t>O) - Proct22:407 2 [0.5 -0.4920] = 0.016 I from standard I normal table -14, 2) unden x -x2-14 12) and corve of -2.406 0 2.406

Here we are dealing with testing of hypothesis. Since we know the standard deviation of populations and we assume to be normal so test statistic follows standard normal distribution.

From the standard normal table we can compute p- value.solution is as follows