Find the 90% confidence interval for the variance of the number of hours spent u
ID: 3262690 • Letter: F
Question
Find the 90% confidence interval for the variance of the number of hours spent using the internet per week if a sample of 38 survey respondents has a standard deviation of 3 hours per week.
Find the 90% confidence interval for the standard deviation of the number of hours spent on vacation per year if a sample of 72 survey respondents has a standard deviation of 4.6 days per year.
The number of M&M's in 16 boxes of M&M's were counted and recorded below. Find the 99% confidence interval for the population standard deviation of the number M&M's per box. Assume the number of M&M's per box is normally distributed. 111 112 126 124 116 134 116 101 103 119 111 105 103 101 125 117
Explanation / Answer
(1-alpha)100% confidence interval for ={ sqrt(n-1)s2/chi-sq( alpha/2 ,n-1),sqrt(n-1)s2/chi-sq(1- alpha/2 ,n-1) }
(first part)90% confidence interval for ={ sqrt((n-1)s2/chi-sq( 0.1/2 ,n-1)),sqrt((n-1)s2/chi-sq(1- 0.1/2 ,n-1)) }
={sqrt((38-1)*3*3/chi-sq( 0.1/2 ,38-1)),sqrt((38-1)*3*3/chi-sq(1- 0.1/2 ,38-1)) }
={sqrt((38-1)*3*3/52.19),sqrt((38-1)*3*3/24.07) }={2.52,3.72
chi-sq(1-0.01/2,37)=24.07, chi-sq(0.1/2,37)=52.19
(second part)
90% confidence interval for ={ sqrt((n-1)s2/chi-sq( 0.1/2 ,n-1)),sqrt((n-1)s2/chi-sq(1- 0.1/2 ,n-1)) }
={sqrt((72-1)*4.6*4.6/chi-sq( 0.1/2 ,72-1)), sqrt((72-1)*4.6*4.6/chi-sq(1- 0.1/2 ,72-1)) }
={sqrt((72-1)*4.6*4.6/91.67), sqrt((72-1)*4.6*4.6/52.6) }={4.05,5.34}
chi-sq(1-0.01/2,71)=52.6, chi-sq(0.1/2,71)=91.67
(third part) using ms-excel s=9.97, n=16
99% confidence interval for ={ sqrt((n-1)s2/chi-sq( 0.01/2 ,n-1)),sqrt((n-1)s2/chi-sq(1- 0.0
1/2 ,n-1)) }
={sqrt((15-1)*9.97*9.97/chi-sq( 0.1/2 ,15-1)), sqrt((15-1)*9.97*9.97/chi-sq(1- 0.1/2 ,15-1)) }
={sqrt((15-1)*9.97*9.97/27.49),sqrt((15-1)*9.97*9.97/6.26) }={7.11,14.91}
chi-sq(1-0.01/2,15)=6.26, chi-sq(0.1/2,15)=27.49
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