Answer to Question #16552 in Statistics and Probability for Angela

Question

normal-shaped distribution has μ = 80 and σ = 15.
a. sketch the distribution of sample means for samples of n = 25 scores from this
population
b. what are the z-score values that form the boundaries for the middle 95% of the distribution of the sample means?
c. compute the z-scores for M = 89 for a sample of n = 25 scores. Is this sample mean in the middle 95% of the distribution? Is this sample mean in the middle 99% of the distribution?
d. compute the z-score for M = 84 for a sample of n = 25 scores. Is this sample mean in the middle 95% of the distribution?

Answer 1

(a) the distributionof sample means for samples of n = 25 scores from this population will be the
normal distribution with mean of 80 and standard deviation of 15/sqrt(25) = 3.
(b) p (z(- x) < z< z( x)) = 0.95 = 2*p(0 < z < z(x)) => z(x) = 1.96 => z = -1.96;
1.96
(c) z(M) = (89 -80)/(15/sqrt(25)) = 3 > 1.96 => this sample mean is not in the middle 95%
of the distribution. For the 99% confidence interval z = -2.575, 2.575. z(M) =
3 does not belong to this interval. So, this sample mean is not in the middle
99% of the distribution.
(d) z(M) = (84 -80)/(15/sqrt(25)) = 4/3 < 1.96 => this sample mean is in the middle 95%
of the distribution.