Construct and interpret a 92% confidence interval for the standard deviation of
ID: 3357863 • Letter: C
Question
Construct and interpret a 92% confidence interval for the standard deviation of age of all passengers.
age 29 0.9167 2 30 25 48 63 39 53 71 47 18 24 26 80 24 50 32 36 37 47 26 42 29 25 25 19 35 28 45 40 30 58 42 45 22 41 48 44 59 60 41 45 42 53 36 58 33 28 17 11 14 36 36 49 36 76 46 47 27 33 36 30 45 27 26 22 47 39 37 64 55 70 36 64 39 38 51 27 33 31 27 31 17 53 4 54 50 27 48 48 49 39 23 38 54 36 36 30 24 28 23 19 64 60 30 50 43 22 60 48 37 35 47 35 22 45 24 49 71 53 19 38 58 23 45 46 25 25 48 49 45 35 40 27 24 55 52 42 55 16 44 51 42 35 35 38 35 38 50 49 46 50 32.5 58 41 42 45 39 49 30 35 42 55 16 51 29 21 30 58 15 30 16 19 18 24 46 54 36 28 65 44 33 37 30 55 47 37 31 23 58 19 64 39 22 65 28.5 45.5 23 29 22 18 17 30 52 47 56 38 22 43 31 45 33 46 36 33 55 54 33 13 18 21 61 48 24 35 30 34 40 35 50 39 56 28 56 56 24 18 24 23 6 45 40 57 32 62 54 43 52 62 67 63 61 48 18 52 39 48 49 17 39 31 40 61 47 35 64 60 60 54 21 55 31 57 45 50 27 50 21 51 21 31 62 36 30 28 30 18 25 34 36 57 18 23 36 28 51 32 19 28 1 4 12 36 34 19 23 26 42 27 24 15 60 40 20 25 36 25 42 42 0.8333 26 22 35 19 44 54 52 37 29 25 45 29 28 29 28 24 8 31 31 22 30 21 8 18 48 28 32 17 29 24 25 18 18 34 54 8 42 34 27 30 23 21 18 40 29 18 36 38 35 38 34 34 16 26 47 21 21 24 24 34 30 52 30 0.6667 24 44 6 28 62 30 7 43 45 24 24 49 48 55 24 32 21 18 20 23 36 54 50 44 29 21 42 63 60 33 17 42 24 47 24 22 32 23 34 24 22 35 45 57 31 26 30 1 3 25 22 17 34 36 24 61 50 42 57 1 31 24 30 40 32 30 46 13 41 19 39 48 70 27 54 39 16 62 32.5 14 2 3 36.5 26 19 28 20 29 39 22 23 29 28 50 19 41 21 19 43 32 34 30 27 2 8 33 36 34 30 28 23 0.8333 3 24 50 19 21 26 25 27 25 18 20 30 59 30 35 40 25 41 25 18.5 14 50 23 28 27 29 27 40 31 30 23 31 12 40 32.5 27 29 2 4 29 0.9167 5 36 33 66 31 26 24 42 13 16 35 16 25 20 18 30 26 40 0.8333 18 26 26 20 24 25 35 18 32 19 4 6 2 17 38 9 11 39 27 26 39 20 26 25 18 24 35 5 9 3 13 5 40 23 38 45 21 23 17 30 23 13 20 32 33 0.75 0.75 5 24 18 40 26 20 18 45 27 22 19 26 22 20 32 21 18 26 6 9 40 32 21 22 20 29 22 22 35 18.5 21 19 18 21 30 18 38 17 17 21 21 21 28 24 16 37 28 24 21 32 29 26 18 20 18 24 36 24 31 31 22 30 70.5 43 35 27 19 30 9 3 36 59 19 17 44 17 22.5 45 22 19 30 29 0.3333 34 28 27 25 24 22 21 17 36.5 36 30 16 1 0.1667 26 33 25 22 36 19 17 42 43 32 19 30 24 23 33 65 24 23 22 18 16 45 39 17 15 47 5 40.5 40.5 18 26 21 9 18 16 48 25 22 16 9 33 41 31 38 9 1 11 10 16 14 40 43 51 32 20 37 28 19 24 17 28 24 20 23.5 41 26 21 45 25 11 27 18 26 23 22 28 28 2 22 43 28 27 42 30 27 25 29 21 20 48 17 34 26 22 33 31 29 4 1 49 33 19 27 23 32 27 20 21 32 17 21 30 21 33 22 4 39 18.5 34.5 44 22 26 4 29 26 1 18 36 25 37 22 26 29 29 22 22 32 34.5 36 39 24 25 45 36 30 20 28 30 26 20.5 27 51 23 32 24 22 29 30.5 35 33 15 35 24 19 55.5 21 24 21 28 25 6 27 34 24 18 22 15 1 20 19 33 12 14 29 28 18 26 21 41 39 21 28.5 22 61 23 22 9 28 42 31 28 32 20 23 20 20 16 31 2 6 3 8 29 1 7 2 16 14 41 21 19 32 0.75 3 26 21 25 22 25 24 28 19 25 18 32 17 24 38 21 10 4 7 2 8 39 22 35 50 47 2 18 41 50 16 25 38.5 14.5 24 21 39 1 24 4 25 20 24.5 29 22 40 21 18 4 10 9 2 40 45 19 30 32 33 23 21 60.5 19 22 31 27 2 29 16 44 25 74 14 24 25 34 0.4167 16 32 30.5 44 25 7 9 29 36 18 63 11.5 40.5 10 36 30 33 28 28 47 18 31 16 31 22 20 14 22 22 32.5 38 51 18 21 47 28.5 21 27 36 27 15 45.5 14.5 26.5 27 29Explanation / Answer
Solution:
From given data first we will find sample size and sample variance,
Sample size = n = 1046
Sample standard variance = 207.75 , Using excel , =VAR(_data_ )
Now we starting procedure of finding CI for standard deviation.
Here crucial step is that we can not find chi square value for given degrees of freedom.
So we can not determine CI for standard deviation.
Done