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)
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.
3 answers
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.
Answer
Answer
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:
3.46<variance<9.29
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:
3.46<variance<9.29
3 answers