Question 1202092: 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 = 60.
population standad deviation = 10.
sample size is 15.
sample mean is 56.
95% two tailed confidence interval has 2.5% on each tail with a critical z-score of plus or minus 1.96.
z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard eror, since you are looking at the mean of a sample rather than an individual score.
standard error = standard deviation / square root of sample size = 10 / sqrt(15) = 2.58199.
the formula becomes z = (56 - 50) / 2.58199 = 2.32379.
area to the right of that z-scorfe is equal to .01007.
since this is less than the critical p-value of .025, the results are significant and the conclusion is that the children are not a random sample from a population of normal children.
the critical z-score confirms this as well since the test z-score is greater than 1.96.
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