How to use TI-84 calculator to solve the problem below? Confidence interval for

Question

How to use TI-84 calculator to solve the problem below? Confidence interval for the standard deviation?

The campus at a certain college has a lake. A student used a Secchi disk to measure the clarity of the lake's water by lowering the disk into the water and measuring the distance below the water surface at which the disk is no longer visible. The following measurements (in inches) were taken on the lake at various points in time over the course of a year.

c. Construct a 95% confidence interval for the standard deviation:

i. x^2a/2 = _____________ , x^21-a/2 = ______________

ii. lower bound = ____________ upper bound = ________________

Explanation / Answer

c.
i.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 5 df are 12.8325 , 0.831
s.d( s^2 )=12.5605
sample size(n)=6
confidence interval for ^2= [ 5 * 157.7662/12.8325 < ^2 < 5 * 157.7662/0.831 ]
= [ 788.8308/12.8325 < ^2 < 788.8308/0.8312 ]
[ 61.4713 < ^2 < 949.0265 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (61.4713) < < sqrt(949.0265), ]
= [ 7.8404 < < 30.8063 ]
^2 = 157.766
lower bound = 7.8404
upper bound = 30.8063

ii.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 5 df are 12.8325 , 0.831
s.d( s^2 )=16.3788
sample size(n)=6
confidence interval for ^2= [ 5 * 268.2651/12.8325 < ^2 < 5 * 268.2651/0.831 ]
= [ 1341.3254/12.8325 < ^2 < 1341.3254/0.8312 ]
[ 104.5257 < ^2 < 1613.7217 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (104.5257) < < sqrt(1613.7217), ]
= [ 10.2238 < < 40.1712 ]

^2 = 268.265
lower bound = 10.2238
upper bound = 40.1712

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