What value of t is the solution of the above equation?
1 answer:
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For a known standard deviation, the confidence interval for sample size = n is
where
x = average
n = sample size
= stad. deviation
z = contant that reflects confidence interval
Let a = x
Let b =
From the given information,
a - b = 0.432 (1)
a + b = 0.52 (2)
Add (1) and (2): 2a = 0.952 => a = 0.476
Subtract (2) from (1): -2b = -0.088 => b = 0.044
Therefore, the confidence interval may be written as
(0.476 - 0.044, 0.476 + 0.044), or as
(0.476
0.044)
where
x = average
n = sample size
z = contant that reflects confidence interval
Let a = x
Let b =
From the given information,
a - b = 0.432 (1)
a + b = 0.52 (2)
Add (1) and (2): 2a = 0.952 => a = 0.476
Subtract (2) from (1): -2b = -0.088 => b = 0.044
Therefore, the confidence interval may be written as
(0.476 - 0.044, 0.476 + 0.044), or as
(0.476