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# HW7 - STAT 500 HW 7 and exercise 8 from HW 6 8 n=8 π=5=...

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STAT 500 HW 7 and exercise 8 from HW 6. 8. n=8; π=5= 0.645, 95% CI; α/2= 0.025=1.960 Cannot use a 1-proportion 2 interval because 8(0.355) = 2.84 < 5 Minitab Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Exact Sample X N Sample p 95% CI P-Value 1 5 8 0.625000 (0.244863, 0.914767) 0.727 HW 7 1. Problem 5.52 a. Confidence interval of 95% t= 1.729 9.10-2.093 * 2.573/√20 = 7.8958 9.10+2.093 * 2.573/√20 = 10.3041 b. The probability plot suggests that the data appear to be a random sample from a normal population distribution. C1 Percent 18 16 14 12 10 8 6 4 2 0 99 95 90 80 70 60 50 40 30 20 10 5 1 Mean 0.508 9.1 StDev 2.573 N 20 AD 0.320 P-Value Probability Plot of C1 Normal - 95% CI c. We are 95% confident that the sample mean reading speed of the fourth graders at the school is between 7.896 and 10.304 minutes

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d. Using a 98% confidence level instead of a 95% confidence level would make the intervals wider. 2. Problem 5.78 _ a. Y= 28.7 b. s= 3.8/√38 = 0.6164 c. 95% confidence interval t= (28.7 – 30)/3.8/√38 t= -1.3/ 0.6164 = -2.11 0.05/2=0.025 t 0.025 = 2.030 for n-1 = 37 (used 35 to be more conservative) Interval: 28.7 – 2.030 * (3.8/√38) = 27.4486 28.7 + 2.030* (3.8/√38) = 29.9513
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