Question

In a test of of garlic for lowering cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured beforeandafter the trmt. The changes in their levels of LDL chol (in mg/dL) have a mean of 5.4 and a sd of 17.5. Construct a 90% confidence interval est. of the mean net change in LDL chol. after the garlic trmt. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?
What is the confidence interval of the mean u?
___Mg/dL < u < ___Mg/dL

Answer 1

\((1-\alpha)\%\) confidence interval for the mean:
\[\left(\hat{\mu}-Z_{\alpha/2}\frac{\hat{\sigma}}{\sqrt n},\,\hat{\mu}+Z_{\alpha/2}\frac{\hat{\sigma}}{\sqrt n}\right)\]
Here you're given \(\hat{\mu}=5.4\), \(\hat{\sigma}=17.5\), and \(n=45\).

You're looking to get 90% confidence, which means \(1-\alpha=0.90\) implies \(\alpha=0.1\), which in turn means \(Z_{\alpha/2}=Z_{0.05}\approx 1.64\).

Answer 2

Thank you for your response. So, 1.64 would then be the the solutions between ___Mg/dL < u < ___Mg/dL?

Answer 3

what*** not 1.64

Answer 4

What would be the confidence intervals be?

Answer 5

Aren't you the person I helped a few days ago? It's the same process as before, with a slight adjustment to the interval's structure.
\[\left(\underbrace{\hat{\mu}-Z_{\alpha/2}\frac{\hat{\sigma}}{\sqrt n}}_\color{red}a,\,\underbrace{\hat{\mu}+Z_{\alpha/2}\frac{\hat{\sigma}}{\sqrt n}}_{\color{blue}b}\right)~~\iff~~P\left(\color{red}a<\mu<\color{blue}b\right)=0.90\]
Your answer should be \(a\) and \(b\). It's a matter of computation at this point.

Answer 6

Yes I am. Sorry, that equation confuses the crap out of me!

Answer 7

How would I compute that on my ti38?

Answer 8

ti83***

Answer 9

Would would E then be? Because to obtain the answer don't I have to add and subtract from X?

Answer 10

What do you mean by "E"?