Question

PROBLEM 1(35pts) You are charged with conducting an investigation in relation to herbicide pollution of IA lakes. Let X denote the act of measuring the level of a certain chemical in any randomly chosen lake. Assume X N(Hr,ax), and that the lakes to be chosen for testing are such that the data collection variables {X^)% can be assumed mutually independent. Let(ux x ) denote the usual estimators of (Ax O,) Data collected on n-50 lakes resulted in (Ax-323. Ở,-69) (a)(12pts) Arrive at the estimate of the 95% 2-sided confidence interval (CI) for each of (Ax,O,). Solution: (b(6pts) Federal law requires that any state having fu, >300 must develop a plan to rectify the situation. Conduct the test Ho : μ,-300 vs. Hı :My > 300 at a significance level α = 0.05 to determine whether or not such a plan will be ordered. Solution: (c)(4pt) In (b) you should have announced H,. Assuming that your t-value was t-2.35, obtain the p-value of the test. Solution: (d)(7pts) A more careful look at the data revealed that for the northern half of IA the results were (4-3 16, σ = 52.01, nl-32) , and for the southern half they were (As = 330, σ2-45.35, n.-18). Test the hypotheses: To this end, note that on p.384 ofthe book we have: Case 1: σ1-02-0:-41-My)-(M-μ)~t Ơ-(n-1)OT + (n2-1)o; is the pooled sample variance, and where n' 2 1s the pooled sample size Solution: NOTE: Compute the numerical values for ỡp ad-It makes my grading easier.] where n,+n,-2 n+ n (e)(6pts) Part (d) assumed that σ-0%-σ . Test the hypotheses: to determine whether this assumption is justified. Solution:

Answer 1

a)
mu_x^ = 323
sigma^= 69

b)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: 300 Ha: μ > 300 standard deviation will be used. Based on the information provided, the significance level is a right-tailed test is te-1.677. 0.05, and the critical value for a The rejection region for this right-tailed test is R-t:t>1.677 The t-statistic is computed as follows: X - 323 -300 70- 460 4 69/V52.357 (4) Decision about the null hypothesis Since it is observed thatt2.357 te1.677, it is then concluded that the null hypothesis is rejected. using the P-value approach. The p-value is p 0.0112, and since p = 0.0112 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean j is greater than 300, at the 0.05 significance level

C)

p-value = 0.0112

D)

The following null and alternative hypotheses need to be tested This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used (2) Rejection Region Based on the information provided, the significance level is α 0.057 and the degrees of freedom are df - 48. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal Hence, it is found that the critical value for this left-tailed test is tc1.677, for a 48 0.05 and df - The rejection region for this left-tailed test is R = {t: t <-1.677) Since it is assumed that the population variances are equal, the t-statistic is computed as follows X1-X2 (n1-1)s+(n2-1)s11 n1 n2 316-330 - -0.955 (32-1)52.01218-1)45.35 1 32T 18 Since it is observed that t-0.955 > te1.677, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p = 0.1722, and since p 0.1722 > 0.05, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μι is less than μ2, at the 0.05 significance level