(a) At 95% confidence interval z- value is 1.96 and t-value is 2.447.
(b) At 80% confidence interval z- value is 1.282 and t-value is 1.440.
(c) At 90% confidence interval z- value is 1.645 and t-value is 1.943.
Given that,
Let's say you randomly choose n=7 measurements from a normal distribution. If you were constructing the following confidence intervals, compare the standard normal z values with the appropriate t values.
We have to find
(a) At 95% confidence interval what is z- value and t-value.
(b) At 80% confidence interval what is z- value and t-value.
(c) At 90% confidence interval what is z- value and t-value.
We know that,
Sample size = n = 7
Degrees of freedom = df = n - 1 = 7 - 1 = 6
(a) At 95% confidence level
α = 1 - 95%
α = 1 - 0.95 =0.05
α/2 = 0.025
Zα/2 = Z0.025 = 1.96
z = 1.96
At 95% confidence level
α= 1 - 95%
α =1 - 0.95 =0.05
α/2 = 0.025
tα/2,df = t0.025,6 = 2.447
t = 2.447
(b) At 80% confidence level
α = 1 - 80%
α = 1 - 0.80 =0.20
α /2 = 0.10
Zα /2 = Z0.10 = 1.282
z = 1.282
At 80% confidence level
α = 1 - 80%
α =1 - 0.80 =0.20
α /2 = 0.10
tα /2,df = t0.10,6 = 1.440
t = 1.440
(c) At 90% confidence level
α = 1 - 90%
α = 1 - 0.90 =0.10
α /2 = 0.05
Zα /2 = Z0.05 = 1.645
z = 1.645
At 90% confidence level
α = 1 - 90%
α =1 - 0.90 =0.10
α /2 = 0.05
tα /2,df = t0.05,6 = 1.943
t = 1.943
Therefore,
(a) At 95% confidence interval z- value is 1.96 and t-value is 2.447.
(b) At 80% confidence interval z- value is 1.282 and t-value is 1.440.
(c) At 90% confidence interval z- value is 1.645 and t-value is 1.943.
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