SOLUTION: We know both the mean and the standard deviation for the population of Total Behavior Problems scores (µ = 50 and σ = 10). Assume that we have a sample of fifteen children who ha
Algebra ->
Probability-and-statistics
-> SOLUTION: We know both the mean and the standard deviation for the population of Total Behavior Problems scores (µ = 50 and σ = 10). Assume that we have a sample of fifteen children who ha
Log On
Question 1202043: We know both the mean and the standard deviation for the population of Total Behavior Problems scores (µ = 50 and σ = 10). Assume that we have a sample of fifteen children who had spent considerable time in a hospital for serious medical reasons, and further suppose that they had a mean score on the Child Behavior Checklist (CBCL) of 56.0. Test the null hypothesis that these fifteen children are a random sample from a population of normal children (i.e., normal with respect to their general level of behavior problems). Use α = 0.05. Set up 95% CI Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! population mean is assumed to be 50
population standard deviation is assumed to be 10.
sample mean is 56.
two tail confidence interval of 95% has critical z-score of plus and minus 1.96.
sample size is 15.
standard error = population standard deviation / square root of sample size = 10 / sqrt(15) = 2.58199 rounded to 5 decimal places.
test z-score = (x-m)/s
x is the sample mean
m is the population mean
s is the standard error.
formula becomes z = (56 - 50) / 2.58199 = 2.32379 rounded to 5 decimal places.
that is considerably more than the critical z-score of 1.96, indicating that the results are significant.
this means that the sample indicates that the differencre in the mean sample versus the population mean indicates that the children sampled most probably don't come from the population of normal children.
