posted by Diandra .
find the 90% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park College if a sample of 24 students has a standard deviation of 2.3 years. Assume the variable is normally distributed. Can someone explain the steps to solve this? Thanks so much.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score for .4500 from mean (Z = 1.645)
90% conf. Interval = mean ± 1.645(SD)
A party host gives a door prize to one guest chosen at random. There are 48 men and 42 women at the party. What is the probability that the prize goes to a woman? Round your answer to 3 decimal places.
That is incorrect. You do not use the Z-score for the mean. This is a Chi-square distribution.
1-.9=.10/2 (for left and right alpha tail) = .05
Find .05 and .95 using the d.f.=23
That equals 34.172 and 13.091
Using the formula for the variance which is ((n-1)s^(2))/(x^(2))
Your equation is an interval, so if you plug each in, your answer should be: