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.
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
To find the 90% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park College, you can follow these steps:
1. Identify the sample size (n) and the sample standard deviation (s). In this case, the sample size is 24 students, and the sample standard deviation is 2.3 years.
2. Determine the degrees of freedom. The degrees of freedom (df) for estimating variance is given by n-1. So, for this problem, df = 24 - 1 = 23.
3. Calculate the chi-square (χ²) values for the confidence level. Since we want a 90% confidence interval, we are interested in the middle 90% of the distribution, leaving 5% in each tail. Therefore, we need to find the critical chi-square values that capture the central 90% of the distribution. Using a chi-square table or calculator, you can find that the lower critical value is 11.689 and the upper critical value is 38.075, for df = 23.
4. Calculate the lower and upper bounds of the confidence interval for the variance. The lower bound (LB) is given by the formula:
LB = (n - 1) * s² / χ² upper
where n is the sample size, s is the sample standard deviation, and χ² upper is the upper critical value of the chi-square distribution.
Plugging in the values from the problem, we have:
LB = (24 - 1) * (2.3)^2 / 38.075
= 1.310
The upper bound (UB) is given by the formula:
UB = (n - 1) * s² / χ² lower
where χ² lower is the lower critical value of the chi-square distribution.
Plugging in the values from the problem, we have:
UB = (24 - 1) * (2.3)^2 / 11.689
= 6.920
Therefore, the 90% confidence interval for the variance is 1.310 to 6.920 (years^2).
5. Calculate the lower and upper bounds of the confidence interval for the standard deviation. The lower bound (LB) is the square root of the lower bound of the variance, and the upper bound (UB) is the square root of the upper bound of the variance.
LB = √1.310 ≈ 1.144
UB = √6.920 ≈ 2.632
Therefore, the 90% confidence interval for the standard deviation is approximately 1.144 to 2.632 (years).
So, the 90% confidence interval for the variance is 1.310 to 6.920 (years^2), and the 90% confidence interval for the standard deviation is approximately 1.144 to 2.632 (years).