Question

ibuted Computea Confidence interval for the true proportion porty certainem (rounded to deci b. find the marginal error of the Confidence interval inquired to de e. We want to test if the true proportion porosity of a certain seam is more than 0.6. (wette enly the convenient sign) 1. Ho: Hy ii. Calculate and write the value of the test statistie: rounded to 2 decimale) W. Find the critical value for this test. Use 1% significance level i What is your conclusion? Une significance level Since the test statistic is less greater) evidence/don't have enough evidence of a certain seam is more than 5 te conclude that the proper d. How large the sample size is necessary if the marginal error of the interval is to be 00187

Answer 1

let, P- the pooportion of porosity of a certain seara, no. of helium porosity of a certain seam. out 01 n 20, Here, p=² @ Here given, that (60%) 9 21:P = 0.4 9% have to computea confidence interval for the true proportion the true proportion porosity of certain 99% seam. given that, significance level (2)=0.01 0.01 (1) 3 The required confidence intervall is, C P-2d/2 P=2016 F Pg Pt 2212 P9 р 2212= Zo.01 2.58 0.6 - 2.58 0.6x04 0.6t 2.58 o 67004 2oo 0.51 0. 69 2 ( Here the margin of error (MoE is MO 2012 pq D CS Samuel Wick CAMSCANNER

MoE = 2.58 200 rimo E = o. 68937 moe seam Claim: IS to be tested, the true proportion porosity of am certain is more than 0.65 Here, the specified value of P = Po - 0.65 Po= |- Po=0.35 i) To test, Hou p = 0.65 Hp: P >0.65 us ii) Test statistic, Zcal P-Po Pogo NNLOAD under Ho 0.6-0.65. 0.65 X 0.35 200 .:2cal -1.48 Hi given that, signifiance level (d) = o.of (1% Critical value = 20.01 Z2 = cs

Page iv) Decision criteria : We reject to at L .l. level of signifi cance Here Z call 7 Zd acal za te - 1.48 K 2-33 accept to at 17 level of significanie - Conclusion: Since, the test statistic is less than the critical value we have enough evidence, +0 conclude that the pro portion porosity of a certain seam is more than 0.55 d @ Here, we have to find sample size (3) Decessary in the marginal error of the 99 % interval is to be 0016. 0.016 2212 pg n 2.58 0.6x0.4 -0.016 1.2639 1,2639 T5 = 78.99 60 n = (7869960)2 6240.375 nr 6240 n ..The required sample size is co] 6240 csl