Question

In desperate need of help:(
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

Answer 1

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

@mathmale

Answer 3

@hartnn @boblovesmath

Answer 4

I am not good with statistics sorry mate, but I can call some help for you.

Answer 5

@satellite73

Answer 6

Thank you!!!

Answer 7

@iPwnBunnies

Answer 8

@mjoxprm360

Answer 9

A 98% CI will have the following form:
\[P\left(a\le\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}\le b\right)=0.98\]
Under the normal distribution assumption, \(a=-2.33\) and \(b=2.33\) are the critical values for a 98% confidence level. Plug in what you know and solve for an interval in terms of \(\mu_1-\mu_2\).

Answer 10

Note: The denominator is written as the standard error, where \(s^2\) replaces \(\sigma^2\) if the latter is not known. In this problem, you're explicitly given the standard deviation, so \(s^2=\sigma^2\) in the formula.