Question

Help with confidence intervals?

Answer 1

Carl conducted an experiment to determine if there is a difference in the mean body temperature between men and women. He found that the mean body temperature for a sample of 100 men was 91.1 with a population standard deviation of 0.52 and the mean body temperature for a sample of 100 women 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 2

@Hero @nincompoop @dan815 @jim_thompson5910

Answer 3

@zepdrix

Answer 4

This question has been asked so many times. Try Googling the first sentence, you'll see what I mean.

Answer 5

I've seen all of the OS posts about it but I I still do not understand.

Answer 6

I just want a simple explanation, please.

Answer 7

I've posted an answer to this one year ago: http://openstudy.com/study#/updates/5378d9a3e4b0f922af0d7ab3

Let me know exactly what you're confused about and I'll be happy to elaborate.

Answer 8

Ohh okay I actually didn't see that one! I recognize the formula, but I'm not sure what to plug in where.

Answer 9

\(\bar{x}_1\) and \(\bar{x}_2\) represent your sample means for the men and women, respectively. For the 100 men you have a mean of \(\bar{x}_1=91.1\), and for the women, the mean is \(\bar{x}_2=97.6\).

\(\mu_1\) and \(\mu_2\) are the hypothesized population means of the men and women. You're looking for a confidence interval for the difference between these two means - this is why you're rearranging the inequality to get something in terms of \(\mu_1-\mu_2\).

In the formula I gave, \(s_1=\sigma_1\) is the population standard deviation of the men, and \(s_2=\sigma_2\) is for the women. These are given: \(\sigma_1=0.52\) and \(\sigma_2=0.45\).

\(n_1\) and \(n_2\) are the sample sizes, in this case both 100.

So, you currently have
\[P\left(a\le\frac{(91.1-97.6)-(\mu_1-\mu_2)}{\sqrt{\dfrac{0.52^2}{100}+\dfrac{0.45^2}{100}}}\le b\right)=0.98\]
Good so far?

Answer 10

One second!

Answer 11

Good so far! Then what?

Answer 12

@sithsandgiggles

Answer 13

Sorry, I was waiting a little over 10 minutes before I had to leave.

So this confidence interval is supposed to have a confidence level of 98%, right? To guarantee that, we want the limits of the interval for the test statistic (that monstrous expression between \(a\) and \(b\)) to match the critical values for this confidence level. You can find out what those critical values are by looking at a \(z\) table.

From the linked post, you know that \(a=-2.33\) and \(b=2.33\) (notice they have the same magnitude but different sign since this is a two-tailed interval).

So you have
\[P\left(-2.33\le\frac{(91.1-97.6)-(\mu_1-\mu_2)}{\sqrt{\dfrac{0.52^2}{100}+\dfrac{0.45^2}{100}}}\le 2.33\right)=0.98\]
The info we're concerned with is contained within the inequality:
\[-2.33\le\frac{(91.1-97.6)-(\mu_1-\mu_2)}{\sqrt{\dfrac{0.52^2}{100}+\dfrac{0.45^2}{100}}}\le 2.33\]
Now you only need to solve for \(\mu_1-\mu_2\), by which I mean manipulate the above so that you have something in the form \(\#_1\le \mu_1-\mu_2\le\#_2\). \(\#_1\) and \(\#_2\) are the lower and upper endpoints of the confidence interval.