Question

Next SIX questions are related to the following data: The following is a random sample of n = 90 undergraduate students' annual textbook expense. 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 1 The point estimate of the population mean is, a $550 b $540 c $530 d $520 2 The variance of the sample is, a 31648 b 27520 c 23930 d 20809 3 To build an interval estimate of the population mean, the standard error of the mean is, a $17.49 b $16.31 c $15.21 d $14.43 4 The margin of error for a 95% interval estimate is, a $30.22 b $32.41 c $34.75 d $37.26 5 The lower and upper ends of the 95% confidence interval are: a $512 $568 b $510 $570 c $508 $572 d $505 $575 6 In the previous question, using a planning value of $150, to obtain a margin of error of ±$20 for a 95% interval estimate the minimum sample size is, a 217 b 235 c 251 d 265.

Answer 1

solution Here the point estimate for Sample mean (x) = x = 1860o = suo Sample variance (89 = {(x-x) = 1852000 ni 90-1 - 20809 Stardand error (SE) = %= 520909 = 15.2056 = 15.2) d-f=n-1=90-1=89 At 95% confidence to/2= 1.987 590

Morgen of error (M. E) = tay SE = 1.987 (15-21 = 30-22 The 95% c-I for el es 2 + M-E = SUO + 30.22 =(509.78, 570.22) = (510, 570) here error (e)=20 24 = tous at a At 95%, Zaje = 1.96 Given o=150 20 thon sample size (n) = (2012 - ) ² = (1.96 (150] = 216.1 ~ 217]