Thread: hypothesis testing

(a) a sample of 14 children from parents without diabetes had a mean fasting glucose level of 82.2mg/dL with sample standard deviation 2.49mg/dL.
give a 95% confidence interval for the underlying standard deviation of fasting glucose level for children from parents without diabetes.
(b)a second sample of 14 children from parents both with type II diabetes had a mean fasting glucose level of 86.1mg/dL with sample standard deviation 2.09mg/dL. perform a hypothesis test to determine if the variability in glucose levels is the same for both groups of children. include a statement of and , the test statistic and its distribution under
(c) based on your answer to (b), perform a hypothesis test to determine if the mean glucose level is higher for children from parents with type II diabetes than from parents without diabetes. include a statement of and , the test statistic and its distribution under

Originally Posted by wik_chick88

(a) a sample of 14 children from parents without diabetes had a mean fasting glucose level of 82.2mg/dL with sample standard deviation 2.49mg/dL.
give a 95% confidence interval for the underlying standard deviation of fasting glucose level for children from parents without diabetes.
(b)a second sample of 14 children from parents both with type II diabetes had a mean fasting glucose level of 86.1mg/dL with sample standard deviation 2.09mg/dL. perform a hypothesis test to determine if the variability in glucose levels is the same for both groups of children. include a statement of and , the test statistic and its distribution under
(c) based on your answer to (b), perform a hypothesis test to determine if the mean glucose level is higher for children from parents with type II diabetes than from parents without diabetes. include a statement of and , the test statistic and its distribution under

What have you tried? What is the specific difficulty you are having here?

CB

for part (a) i think i have figured it out...
= ?
n = 14
s = 2.49

from the chi-square-distribution table i got
= 5.01
= 24.7

so a 95% confidence interval for is

= (3.263, 16.088)

so a 95% confidence interval for is
(1.806, 4.011)

firstly, have i done this correctly? i was referring to the poorly set out example in our textbook! also, can u please explain why you use the values = 0.025 and 0.975 when referring to the chi-square-distribution table?

You want .95 probability between any two percentiles
so you can use those two, the .025 and the.975 percentiles, which most do
That puts equal probability in each tail
Or you can use one sided CIs where all the .05 is in one tail
or use .02 and .97 etc.........

And these are CIs for and

is the underlying distribution

In part (b) you want to perfrom an F test comparing the two POPULATION variances
Then in (c) you will a t test comparing the two means.
NOW if you come away from (b) saying that the two st deviations are equal you
will perform a pooled t test, where you pool the sample st deviations
IF not you will use the non pooled test where you need to approximate the nasty degrees of freedom using satterwhite's approximation
Student's t-test - Wikipedia, the free encyclopedia
also http://www.bsos.umd.edu/socy/alan/ha..._two_means.pdf