Question
In: Statistics and Probability
Data are collected from 12 randomly selected infants undergoing nutritional therapy following premature birth. The data...
Data are collected from 12 randomly selected infants undergoing nutritional therapy following premature birth. The data represent the percent gain in birthweight after 8 weeks of therapy. Generate a 95% CI for the mean percent gain in weight.
23% 32% 48% 4% 34% 20% 11% 24% 40% 69% 17% 6%
Solutions
Expert Solution
Solution:
sample mean = (23+32+48+4+34+20+11+24+40+69+17+6)/12 = 328/12 =
27.33
Sample standard deviation = sqrt(summation(Xi-Xbar)^2/(n-1))
|
X |
(Xi-Xbar) |
(Xi-Xbar)^2 |
|
23 |
-4.33333333333333 |
18.7777777777778 |
|
32 |
4.66666666666667 |
21.7777777777778 |
|
48 |
20.6666666666667 |
427.111111111111 |
|
4 |
-23.3333333333333 |
544.444444444444 |
|
34 |
6.66666666666667 |
44.4444444444445 |
|
20 |
-7.33333333333333 |
53.7777777777778 |
|
11 |
-16.3333333333333 |
266.777777777778 |
|
24 |
-3.33333333333333 |
11.1111111111111 |
|
40 |
12.6666666666667 |
160.444444444444 |
|
69 |
41.6666666666667 |
1736.11111111111 |
|
17 |
-10.3333333333333 |
106.777777777778 |
|
6 |
-21.3333333333333 |
455.111111111111 |
|
Summation |
3846.66666666667 |
Standard devaition = sqrt(3846.66/11) = 18.7
95% confidence interval is
27.33 +/- talpha/2 * SD/sqrt(n)
27.33 +/-2.20*18.7/sqrt(12)
27.33 +/-11.87
so 95% confidence interval is
15.45% TO 39.21
orchestra answered 1 year agoRelated Solutions
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