Given the confidence interval for a mean of (74.5738,77.4262), from a sample of size 34 with a population standard deviation of σ=3.6, find the following:
Margin of Error(ME)=
Standard Error(SE)=
Zc=
what was the confidence level for this confidence interval?=
(Show your work please)
confidence interval is
lower limit = 74.5738
upper limit= 77.4262
sample mean = (lower limit+upper limit)/2= (
74.5738 + 77.4262 ) / 2
= 76
margin of error = (upper limit-lower limit)/2=
( 77.4262 - 74.5738
) / 2 = 1.4262
Standard Error ,
SE = σ/√n = 3.6000 / √
34 = 0.6174
margin of error ,E = Z(*σ/√n )
Z (α/2)= E*√n /σ = 1.426 * √
34 / 3.600
= 2.310
α/2 from critical value 2.310
is
α/2= 0.0104
α= 0.0209
confidence level = 1-α= 1-
0.0209 = 0.9791 or
97.91%
Given the confidence interval for a mean of (74.5738,77.4262), from a sample of size 34 with...
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