Question 1184273: Assume that in a certain district the mean systolic blood pressure of persons aged 20 to 40
is 130 mm Hg with a standard deviation of 10 mm Hg. A random sample of 64 persons
aged 20 to 40 from village X of the same district has a mean systolic blood pressure of 132
mm Hg. Does the mean systolic blood pressure of the dwellers of the village (aged 20 to
40) differ from that of the inhabitants of the district (aged 20 to 40) in general, at the 5%
level of significance?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean = 130
population standard deviation = 10
sample size = 64
sample mean = 132
two-tailed critical z-score at 5% significance level = plus or minus 1.96.
standard error = population standard deviation / square root of sample size.
standard error = 10 / sqrt(64) = 10 / 8 = 1.25
sample z-score = (sample mean minus population mean) / standard error
sample z-score = (132 - 130) / 1.25 = 1.6
the results of the test are not significant, since the absolute value of the test z-score is not greater than the absolute value of the critical z-score.
in other words, the different mean of the sample is more then likely due to variations in the mean of samples size 64, rather than something out of the ordinary, as determined by the significance level used for the test.
to find the critical raw scores, use the z-score formula and do the following:
when z = -1.96:
-1.96 = (x - 130) / 1.25.
solve for x to get:
x = -1.96 * 1.25 + 130 = 127.55
when z = 1.96:
1.96 = (x - 130) / 1.25
solve for x to get:
x = 1.96 * 1.25 + 130 = 132.45
the sample mean of 132 is between these two critical values, therefore the test is not considered to be significant, leading to the conclusion that there is no significant difference between the mean of the population and the mean of the sample.
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