Question

1. The melting point of each 16 samples of a certain brand of hydrogenated vegetable oil was = 94.32. Assume that the distribution of melting point is normal determined, resulting with ơ-1.20. (a) Test llo : μ-95 versus lla : μ 95 using the significance level .01 test. (b) If level a0 test is used, what is B(94), the probability of type error when μ=94? (c) What value of n is necessary to ensure that β(94)-0.10 when α-001?

Answer 1

(a) Step1: The null and altenative hypotheses are Ho : μ= 95 (The population mean is equal to 95) H1 : μ 95 (The population mean is not equal to 95) The test is two-tailed test, because we are testing the population mean is equal to 95 Step2: Decision Rule The significance level is 0.01. The sign in the alternative hypothesis indicates that the test is two-tailed with the rejection region in the right-tail andleft-tail of the sampling distribution curve of x. The critical value of z, obtained from the normal table for 0.005 area in the right tail, and in the left tail is given below. The table value of the z test stati sti c is zoos-t2.575 Reject Ho if Z > 2.575 otherwise do not reject Ho Step3: The summary stati stics: n-16,I-94.32, ơ-1.20 The formula for calculating thez- test stati stic is 94.32-95 16 -0.68 4 0.68 0.3 --2.266666667 2.27 z Rs-2.27 Step4: Since, the test statistic is zo >os 2.575, we reject the null hypothesis Ho. Reject the null hypothesis Ho-. Step5: The P- value approach 2P(z >4)-2(1-P(z <%)) 2(1-P(z < 2.2667) 2(1-0.988294702) 2(0011705298) 0.023410596 Since, on using Excel2010 function 0.0234 normsdist(z) 0.01. We do not reiect the hypothesis H P-value0.0234 Step6: Interpretati on Since, the z-test stati stic isol 227 o2.575. We do not reject the null hypothesis Ho There is not sufficient evidence to reject the null hypothesis Ho Hence, we can conclude that the population mean is equal to 95

(b) The populati on mean is 95 The real value mean is μ= 94 The standard deviation is σ 12. The sample size is n 16 =95-94 The formula for calculating the Type I error of the test is Type II error-Φ| z"12 00121 (25758-豐)-φ(-25758- 1(4) → Ф| 1.2 → Φ(25758-3333333)-$(-2.5758-3333333) Φ(-0.75753)-Φ(-5.909133) 0.224365-0.0000000017 0.2243652000 β~ [0.22441 Therefore, the power of the testis 0.2244

c) The level of significance is 0.01 The power of the test is 0.10. The formula for calcualting the sample size is (zoourz +Zo) (12) 0.01/20 20005 +2010)2 (1 2)2 (2.5758+1.28155) (1.2) 14.87938717) (144) 21.42631752 n-22 Therefore, the necessary sample size is approxim ately n -