Thread: Confidence Interval for unknown sigma

1. Confidence Interval for unknown sigma

Suppose the height of college students can be modeled by the normal distribution. We believe the mean of the normal distribution is 1.76 meters. We randomly sample 18 students and find an average of 1.69 meters and a sample standard deviation of 12 cm (=.12 meters)

Find an 80% confidence interval for mu.

Why isn't mu given in this interval.

Would this be used for a t-dist or for a z-score?
Since there is a sample its a t-dist correct? But how do you compute the sigma for the population in order to find the CI?

2. a 2-sided CI for the pop mean is

$\displaystyle \bar X\pm t_{n-1,\alpha/2}S/\sqrt{n}$

3. CI Assistance

Hey, (first time posting)

You are right about it being a t distribution because (based on what you got) variance is unknown.

If you just want to find the CI interval for this problem, then the value sigma is unnecessary.

The formula for a two sided CI for a t-distribution was given above. All you need to do now is figure out what each variable in the formula means in terms of your problem and substitute into the formula.

(Hope this isn't too vague.)