SOLUTION: This question was answered, but the answer confused me even more - So I will enter the Table we are to use also then maybe you folks can help unconfuse me.
Problem: If we wish to
Algebra ->
Probability-and-statistics
-> SOLUTION: This question was answered, but the answer confused me even more - So I will enter the Table we are to use also then maybe you folks can help unconfuse me.
Problem: If we wish to
Log On
Question 662459: This question was answered, but the answer confused me even more - So I will enter the Table we are to use also then maybe you folks can help unconfuse me.
Problem: If we wish to have a 99% confidence interval, what would be the value of the confidence coefficient, z(a/2)? (Use Table 4B)
Table 4B:
a.....0.25...0.20...0.10...0.05...0.02...0.01
z(a/2..1.15...1.28..1.65...1.96...2.33...2.58
1-a....0.75...0.80..0.90...0.95...0.98...0.99 Answer by ewatrrr(24785) (Show Source):
Hi,
certainly can understand Your diligence... translating .995 to a z-value gives 2.58
Table 4B:
a.....0.25...0.20...0.10...0.05...0.02...0.01 Percents listed as decimals
z(a/2)..1.15...1.28..1.65...1.96...2.33...
1-a....0.75...0.80..0.90...0.95...0.98...0.99 Percents listed as decimals
1-a/2....0.875...0.90..0.95...0.975...0.99...0.995 and NORMSDIST(.995) = 2.58
