Statistics

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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.

  • Statistics -

    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)

  • Statistics -

    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

  • Statistics -

    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

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