If one were to use a standard normal distribution instead of Student's t distribution to calculate a confidence interval, what would be the "critical normal value" used as the coefficient for the s.e.m for a 90% CI?
Solution :
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2
= Z 0.05 = 1.645
Critical normal value = 1.645
If one were to use a standard normal distribution instead of Student's t distribution to calculate...
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