Carla wants to determine if there is a difference in the mean body temperature between men and women. She knows that mean body temperature for men was 97.1 with a population standard deviation of 0.51 and the mean body temperature for women is 98.2 with a population standard deviation of 0.57. Assuming the population of body temperatures for men and women is normally distributed, calculate the 95% 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 contextof theproblem
@abb0t i cant get anybody to help me can you? i wanna know how to solve these
@thomaster help?
This question's been asked at least two other times, and the same problem shows up. You need to have the sample size in order to construct the interval and margin of error.
thats the problem this is all it gives me
The best you can do is construct the interval in terms of an unknown sample size \(n\) (if the same number of men and women are used) or \(n_1\) and \(n_2\) (if they're not the same number). The interval for a \((1-\alpha)100\%\) confidence level has the form \[\left((\bar{x}_1-\bar{x}_2)-Z_{\alpha/2}\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}},~(\bar{x}_1-\bar{x}_2)+Z_{\alpha/2}\sqrt{\frac{{\sigma_1}^2}{{n_1}^2}+\frac{{\sigma_2}^2}{{n_2}^2}}\right)\]
by the way thx for reading this ur the first person to help me and i have been trying to get this question for 2 days also i think thats all i need
No problem. The last person to ask this didn't seem to understand that the given info is incomplete. By the way, the margin of error is the \(Z_{\alpha/2}\sqrt{\cdots}\) term.
what would the interval of theconfidence 95% interval
...seriously? I just said this question is unanswerable.
i was wandering if the way to solve that interval all i need is to set up the equation if it was solvable
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