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Unformatted text preview: Q = 1-p =1- 0.75 =0.27 n = 0.73 x 0.27 (1.96 over 0.03) ^2 = 841.310 n=842 P =0.5 n= pq (2/2 over E) ^2 = 0.5 x 0.5 ( 1.96 over 0.03) ^2 = 1067.111 N= 1067 6. n=g = n-1 = 8 X = x over n = 209.7444 over s = (x-x) ^2 over n-1 = x^2 - 1 over n (x) ^2 over n-1 = 1.673403T = (209.7444 -208) over 1.673403 √ ❑ = 3.1273 = 3.127 Test statistic 3.127 a= 0.01 Critical value 2.8963.127 > 2.896 t> 01;8Ho is rejected There is sufficient evidence at 0.01 levels to conclude that the weight is greater than 208 grams. 7. Proportion of children who got free lunch p= 156 over 300 = 0.52 Ho the proportion of children who got lunch free is 0.593 H1 the proportion of children who got lunch free is less than 0.593 z= p-p over √ ❑ over n = 0.52 - 0.593 over √ ❑The p value for the test is P=P( Z < -2.574) = 0.0051Since P value is less than the significaNT level a= 0.01 we reject the null hypothesis Hence, we conclude that the proportion of children who got lunch free is less than 0.593 8. n=[ z/2 over E] = [ 258(2.6) over 0.5]^2 n= 179.98 or 180...
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