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I just need help with questions 5 and 6
370 300 630 530 Next SIX questions are related to the following data: The following is a random sample of n = 90 undergraduat
d $37.26 The lower and upper ends of the 95% confidence interval are: a $512 $568 $510 $570 $508 $572 Dd $505 $575 2 6 13 74
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Answer #1

610,600,300,420,520,470,430,520,400,370,730,480,450,500,650,370,540,330,690,550,450,450,750,750,660,700,300,770,760,390,680,450,590,630,530,700,580,390,330,320,350,490,310,320,780,590,370,470,760,550,630,450,640,620,520,440,720,660,440,770,380,450,800,720,370,550,460,560,590,350,790,700,550,650,750,400,490,690,610,730,560,420,410,720,650,450,540,590,310,370

Solution: by using above the samples : n = 90, df = 89
  
1. option b. $540.

The point estimate of the population mean is 540
=> mean = sum of terms/number of terms
= 48600/90
= 540

5. b. $510 $570

Margin of error = t*s/sqrt(n) = 1.987*144.2532/sqrt(90) = 30.21

95% Confidence interval are = 540 +/- 30.21 = (510, 570)

6. option A. 217

n = (Z*s/E)^2 = (1.96*150/20)^2 = 216.09 = 217

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