Instead, you should be using the t distribution with 39 degrees of freedom.

[joemama69]

Homework Statement

Measure of Contaminatino has a normal distribution w/ unknown mean and unknown variance. Random sample of n=40 provides sample mean = 28.30 and sample variance = 17.38

1)Find Upper 95% Confidence Interval

2)Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Homework Equations

The Attempt at a Solution

1)...95% CI... therefore z(1.645) = 95%

(xbar)+z sqrt(sigma^2 / n) = 28.30 + 1.645 sqrt(17.38 / 40) = 28.30 + 1.0843=29.38

(0 < u < 29.38)


2) Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Since n = 40 we do not to use T-Tables and can just compute the P-Value from Z-Table.

Z=(x-u)/sqrt(sigma^2/n) = (28.30 - 25.50)/sqrt(17.38 / 40) = z(4.25) = 1.00 ?

Since it us upper tailed P-Value = 1-z(4.25) = 1-1 = 0 ?


Where did I go wrong... Based on my 95% CI, Shouldnt I get a little less than 95% for a value lessthan 29.38?

 

[Ray Vickson]

Homework Statement

Measure of Contaminatino has a normal distribution w/ unknown mean and unknown variance. Random sample of n=40 provides sample mean = 28.30 and sample variance = 17.38

1)Find Upper 95% Confidence Interval

2)Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Homework Equations

The Attempt at a Solution

1)...95% CI... therefore z(1.645) = 95%

(xbar)+z sqrt(sigma^2 / n) = 28.30 + 1.645 sqrt(17.38 / 40) = 28.30 + 1.0843=29.38

(0 < u < 29.38)


2) Test Null Hyp = 25.5 vs Alt Hyp = 28.30... Upper Tailed

Since n = 40 we do not to use T-Tables and can just compute the P-Value from Z-Table.

Z=(x-u)/sqrt(sigma^2/n) = (28.30 - 25.50)/sqrt(17.38 / 40) = z(4.25) = 1.00 ?

Since it us upper tailed P-Value = 1-z(4.25) = 1-1 = 0 ?


Where did I go wrong... Based on my 95% CI, Shouldnt I get a little less than 95% for a value lessthan 29.38?

The questions make no sense: 95% confidence interval of WHAT? The mean? The variance? The actual level of contamination?

If you do want a confidence interval for the mean, you are using the wrong distribution: when you have to estimate both the mean and the variance from the data, the normal distribution does not apply. (It would apply if you knew the variance exactly and only had to estimate the mean from the data.)