ADM 2304 Lecture Notes - Lecture 7: Histogram, Box Plot, Confidence Interval
19 Nov 2015
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So z=(0. 044-0. 0678)/ ((0. 0678*o. 9322)/1000))=-2. 9936 (zcalc)> (zcrit=z =1. 645) so reject the ho. there is sufficient evidence that supported for the greens dropped: since me=0. 01, and the me almost equal to 2se(p-hat) And it is a 99% two sided interval. Since true proportion is 0. 0678, so (0. 0678*0. 9322)/n=64. 76. As a result, we assume the sample size should be 4194: since np=17*0. 3962=6. 7354<10 nq=17*(1-0. 3962)=17*0. 6038=10. 2646>10. Since np is not greater than 10, we have to use binomial to solve. Since this question doesn"t meet the assumptions for normal distribution. Therefore, we can"t infer anything about the national level of support for the conservatives. Variable n n* mean se mean stdev minimum q1 median q3 maximum avginc 268 0 47994 815 13341 22511 39813 46433 55273 135608 b. Variable n mean stdev se mean 90% ci. It is the same answer in the minitab. Histogram of c4 y c n e u q e r.
