Question

(t point) The Wolf River in Tennessee flows pastan abandoned site once used by the pesticide industry for dumping wastes including chordane, aldrin, and diedrin. These highly toxic organic compounds can cause various cancers and birth defects. The standard method to test whether these poisons are present in a river is to take samples attenth depth (c. depths that are six-tenths of the way from the surface to the bottom. The organic compounds in question are denser than water, and their molecules tend to stick to particles of sediment. Both of these facts suggest that you'd be likely to find higher concentrations near the bottom than near mid-depth. In this question we will be focusing on 6 measurements of aldrin concentration at both bottom and mid-depth. Suppose the aldrin data actually had been collected so that the data were paired, that is, the bottom and mid- depth measurements in the same row were taken at the same time and in the same place. The data are given in the following Bottom:3.8, 4.8, 48, 5.1, 5.8,7 Mid-Depth: 32, 3.8.4.2, 4.4, 4.5,53 You may assume any necessary conditions for inference. Unless otherwise stated, give your answers to three decimal places. 1. Construct a 90% confidence interval for the average of the difference in aldrin concentration between bottom and mid- depth use bottom mid-depth. Give your answers to three decimal places - 2. Conduct a hypothesis test to determine whether the average aldri concentration is higher at the bottom than at mid-depth. 3. What is the parameter of interest? OLP 5 7 OL WP- b. What is the correct null value for this? c. What sign should spear in the rative hypothesi? OL- OCLC d. The best statistic for this festist The value for this testis p-value < 0.001, 1. Select the propriate decision for this hypothesis 0.01) OLH, O. Falto reject I. O Acopt Select the proprietarpretation for this hypothesis 0.01) CL We do not have statistically significant evidence to the concentration at the bomomis greer than the world concentration at midden have a significant evidence to get there is no erence in the age 2 A MacBook Pro 30 7 2 3 5 6 7 8 9 W E R T Y U А S D F G H K Z C р

Answer 1

Let denote the average concentration at bottom and mid-depth respectively.

111, 112

1.

2.

Here, 21 = 5.217,81 = 1.0852, n1 = 6,72 = 4.233, S2 = 0.7062, n2 = 6 and associated degrees of freedom df= = 9 Ĉ + ni -1 n2-1 The Standard error is 81 1.092 SE= + + 0.712 6 0.529 ni n2 6 The critical value is tcritical = t10.0 2,9 =t0.05,9 = 1.8331131.833 The margin of error is 1.092 0.712 1.092 0.712 ME=t1 0.9 ,9 * + =t0.050,9 * + 1.833* + 0.968933 0.969 ni n2 6 6 6 6 a 90 percent confidence interval for the differnce in population means (Mi - U2 ) is si S = či - 72 + *t 1-0.900 ni n2 = 21 - 22 + ME = (5.217 – 4.233) +0.968933 = 0.983 +0.968933 (0.014400, 1.952266) (0.014, 1.952)

Hypotheses : To test null hypothesis HM1-M2 = 0 against alternative hypothesis H1 H1-42 > 0 This is a one tailed right sided test as the alternative hypothesis contains '>' sign. Here, C1 = 5.22, S1 = 1.09, n1 = 6,72 = 4.23, 82 = 0.71, n2 6 The test statistic can be written as: (21 02-0 ta 5 + $ 112 V m which under H, follows at distribution with 8 df. where + 1.092 + 0.71 Ti df= 8 (1.092) 0 1 (719) 61 + + -1 T2-1 and standard error for the difference between means 87 82 1.092 SE= + + 0.712 6 = 0.528573 ni n2 6 Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if tstat > t0.01,df 0.4 - 0.3- 20.2- 30.1- 0.0- -2 t Now, Value of the test statistic : The value of the test statistic is tstat = (5.22 - 4.23) 1.09 + 0.71 = 1.860357 1.860 6 associated degrees of freedom = 8 The critical value = t critical = t0.01,8 =2.896459 2.896 P-value = P(ts > tstat) = P(t: > 1.860357)=1- P(t< 1.860357) = 1-0.950062 = 0.049938 0.0499 Conclusion : Since p-value > 0.01 and tstat > tcritical = 2.896, so We fail to reject H, at 0.01 level of significance Hence, we can conclude that mean for first population is not significantly greater than the mean for second population.

g) ans-> i)