# Lab 3 - 1. True Margin of error is E=z * 0 over √ ❑ as n...

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1.True, Margin of error is E=z * 0 overas n increases e decreases2.True because E = z * o overTo determine the sample size you need to estimatethe parameter, you must know the margin of error.3.False The null hypothesis has an equality sign. Alternative is opposite to null. It have≠,>,<signs4.False If p value is less thanδ ,the nullhypothesisisrejected.5.True The confidence interval can be used to test or claim a hypothesis.2. za / 2 o overa=1 - c = 1 - 0.90= 0.10 o/2 = 0.20/2 =0.05a/2 = 1.645 x - a/2 o over121.60 - 1.645 6.36 over121.60 - 1.7437 < u < 121.60 + 1.7437119.8563 < u < 123.3437The 90% confidence interval for true mean is ( 119.8563, 123.3437)3. 93 + 83+ 76+ 92+77 + 81 + 78 +100 + 78 + 76 +75 over 11= 909 over 11 = 82.6364[x¿22¿Σ x¿over n=[ 75837 - 909 ^2 over 10=The value of true mean with 95% confidence82.6346 0.025 x 8.4885 over82.6346 + 1.96 x 8.4885 over(77.62, 8765)4. N * p = 1500 * 0.42 =630=confidence interval = ( X-Z * s/(630 - 1.96 * 19.12/= (629.0324 , 630.9676)
5. Confidence level is 95%x= 1- 0.95= 0.05Critical value2/2 =196n= pq ( 2/2 over E)
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Term
Fall
Professor
Amitesh Maiti
Tags
Statistical hypothesis testing, true mean

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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