Question

MEDAL?!
How do you find the confidence interval?

Answer 1

@ganeshie8

Answer 2

1.Identify a sample statistic. Choose the statistic (e.g, sample mean, sample proportion) that you will use to estimate a population parameter.
2.Select a confidence level. ...
3.Find the margin of error. ...
4.Specify the confidence interval.

Answer 3

Is there a formula for the confidence interval? & my problem only shows the sample and standard deviation...

Answer 4

Za/2 * σ/√(n). Za/2..... a = confidence level, σ = standard deviation, and n = sample size tell me if your still stuck

Answer 5

What does the Z stand for?

Answer 6

on this part, σ/√(n). Za/2 is there a mutiplication symbol in between the two fractions?

Answer 7

the mean i think not completely sure

Answer 8

and there's not a multiplication symbol in between it

Answer 9

then whats between it?

Answer 10

The confidence interval for a \((1-\alpha)\%\) confidence level is given by
\[\large \bigg(\theta_0-Z_{\alpha/2}\frac{\sigma}{\sqrt n},~\theta_0+Z_{\alpha/2}\frac{\sigma}{\sqrt n}\bigg)\]
\(\theta_0\) is the measured statistic, \(Z_{\alpha/2}\) is the cuttoff/critical value, and \(\dfrac{\sigma}{\sqrt n}\) is the standard error. \(\sigma\) is the population standard deviation (if known) or can be estimated by a sample standard deviation. \(n\) is the sample size.

The cutoff value depends on the test you wish to use, and \(\theta_0\) depends on the statistic you wish to estimate.

Answer 11

If you have an actual question to work with, I think it'd be easier to see what goes where.

Answer 12

Would you like me to post it?

Answer 13

Only if you need help with it :P

Answer 14

Okay c: its:

Carl conducted an experiment to determine if the there is a difference in mean body temperature between men and women. He found that the mean body temperature for men in the sample was 91.1 with a population standard deviation of 0.52 and the mean body temperature for women in the sample was 97.6 with a population standard deviation of 0.45.

Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, explain what these confidence intervals mean in the context of the problem.

Answer 15

In this context:
\[\alpha=0.02~~~\text{(because you want a 98% confidence level)}\\
\begin{matrix}\bar{x}_1=91.1&&\bar{x}_2=97.6\\\sigma_1=0.52&&\sigma_2=0.45\end{matrix}\]
The test is supposed to examine if there's a difference between the means for men and women, which means you want to estimate the difference between the population means: \(\mu_1-\mu_2\).

The estimator for this statistic is the difference between the sample means: \(\bar{x}_1-\bar{x}_2\).

The critical values would be \(\large \pm Z_{0.02/2}=\pm Z_{0.01}=2.33\) (refer to a table).

The standard error for this statistic is
\[\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}}\]
So, the 98% CI will have the following form:
\[\left((\bar{x}_1-\bar{x}_2)-2.33\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}},~(\bar{x}_1-\bar{x}_2)+2.33\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}}\right)\]
You can plug in nearly everything, but you're missing the sample sizes...

Answer 16

so i wouldnt be able to get a definite answer without the sample sizes?

Answer 17

Right

Answer 18

thanks C: !

Answer 19

yw