Confidence Interval Calculation

[MMCS]

See attached for problem

Working:

Mean: 1203.26
Standard Deviation: 7.047

for 99% confidence interval, level of significance = 0.01
Therefore

0.5 - 0.01/2 = 0.45
Reading 0.45 from standard distribution table i get a value of 1.65

therefore confidence interval should be

1203.26 +/- (1.65*7.047)/√15

1203.26 +/- 3

however using this website to check my answer:

http://www.mccallum-layton.co.uk/tools/statistic-calculators/confidence-interval-for-mean-calculator/

It gives me a value of 5 (approx)

Thanks

 

[Ray Vickson]

See attached for problem

Working:

Mean: 1203.26
Standard Deviation: 7.047

for 99% confidence interval, level of significance = 0.01
Therefore

0.5 - 0.01/2 = 0.45
Reading 0.45 from standard distribution table i get a value of 1.65

therefore confidence interval should be

1203.26 +/- (1.65*7.047)/√15

1203.26 +/- 3

however using this website to check my answer:

http://www.mccallum-layton.co.uk/tools/statistic-calculators/confidence-interval-for-mean-calculator/

It gives me a value of 5 (approx)

Thanks

For a (symmetric) 99% confidence on a quantity X, you want a 0.5% chance that X is below below the lower limit and a 0.5% chance that X is above the upper limit. In other words, if U is the upper limit you want Pr{ X ≤ U} = 1-0.005 = 0.995. What is the 99.5 percentile on the standard normal distribution?

 

[MMCS]

For a (symmetric) 99% confidence on a quantity X, you want a 0.5% chance that X is below below the lower limit and a 0.5% chance that X is above the upper limit. In other words, if U is the upper limit you want Pr{ X ≤ U} = 1-0.005 = 0.995. What is the 99.5 percentile on the standard normal distribution?

Thanks for your reply, I'm not sure where the 0.005 value is from?

 

[Ray Vickson]

Thanks for your reply, I'm not sure where the 0.005 value is from?

What is meant by x%?