Posted by Rob on Monday, March 28, 2011 at 11:43am.
An obstetrician maintains that preterm babies (gestation less than 37 weeks) have a higher variability of birth weight than do full-term babies (gestation 37-41 weeks). According to the National Vital Statistics Report, Vol. 48, No. 3, the birth weights of full time babies are normally distributed with a standard deviation of 505.6 grams. A random sample of 41 preterm babies results in a standard deviation of 840 grams. Test the obstetrician's claim that the variability in birth weight for preterm babies is more than that of full-term babies, at the level of significance 0.01.
Any help on figuring this out would be greatly appreciated. I need the mean to figure this out, but I feel like the question is leaving something short. Thanks for your time
Stats. - MathGuru, Thursday, March 31, 2011 at 12:24pm
Null and alternative hypotheses:
Ho: The population standard deviation is = 505.6
Ha: The population standard deviation is > 505.6
sample standard deviation = 840 (Note: standard deviation is the square root of the variance; therefore, variance is standard deviation squared).
level of significance = 0.01
n = 41
The test statistic used in making a decision is the chi-square.
Here's the formula:
chi-square = (sample size - 1)(sample variance)/(value specified in the null hypothesis)
Note: Convert both standard deviations to variances (505.6 and 840) when working with the formula.
Degrees of freedom is equal to n - 1, which is 40.
Checking a chi-square table using alpha = 0.01 with 40 degrees of freedom, find the critical value. If the critical value exceeds the test statistic, the null is rejected and there is a difference. If the critical value does not exceed the test statistic, then the null cannot be rejected and there is not enough evidence to support the claim that the population standard deviation is greater than 505.6 grams.
I hope this will help get you started.
Stats. - MathGuru, Thursday, March 31, 2011 at 12:35pm
If the test statistic exceeds the critical value from the chi-square table, then the null is rejected. Also, if the test statistic does not exceed the critical value from the table, then the null cannot be rejected.
The above corrects the last few sentences. Sorry for any confusion.
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