In: Statistics and Probability
Need a second opinion on this Stats question.. thanks
The following data represents the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data is approximately normal with a sample standard deviation of 10.00 weeks. Construct and interpret a 95% confidence interval for the population standard deviation of the age at which babies first crawl [52 30 44 35 39 26 47 37 56 26 39 28]
Solution :
Given that,
s = 10
s2 = 100
n = 12
Degrees of freedom = df = n - 1 = 11
At 95% confidence level the
2 value is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
1 -
/ 2 = 1 - 0.025 = 0.975
2L
=
2
/2,df
= 21.92
2R
=
21 -
/2,df = 3.816
The 95% confidence interval for
is,
(n
- 1)s2 /
2
/2
<
<
(n - 1)s2 /
21 -
/2
12
* 100 / 21.92 <
<
12 * 100 / 3.816
7.08 <
< 16.98
(7.08 , 16.98 )