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
Null and alternative hypotheses:
Ho: The population standard deviation is = 505.6
Ha: The population standard deviation is > 505.6
Data given:
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.
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.
To test the obstetrician's claim, we need to compare the variability in birth weight for preterm babies to that of full-term babies. This can be done by conducting a hypothesis test.
Let's start by setting up the null and alternative hypotheses:
Null Hypothesis (H0): Variability in birth weight for preterm babies is equal to or less than that of full-term babies.
Alternative Hypothesis (Ha): Variability in birth weight for preterm babies is greater than that of full-term babies.
Since we are testing the claim that the variability for preterm babies is more than that of full-term babies, this is a one-tailed test (specifically, a right-tailed test).
To conduct the test, we will use the F-test statistic. The F-test compares the variances of two populations.
The formula for the F-test statistic is:
F = (s1^2) / (s2^2)
where s1 is the standard deviation of the first sample (preterm babies) and s2 is the standard deviation of the second sample (full-term babies).
In this case, s1 = 840 grams (from the sample of 41 preterm babies) and s2 = 505.6 grams (from the National Vital Statistics Report).
Now, let's calculate the F-test statistic:
F = (840^2) / (505.6^2) = 1.782
Next, we need to determine the critical value for the F-distribution. This is based on the significance level (α) and the degrees of freedom for both samples.
Degrees of freedom for the numerator (preterm babies) = sample size - 1 = 41 - 1 = 40
Degrees of freedom for the denominator (full-term babies) = sample size - 1 = 1 - 1 = 0
Since the denominator degrees of freedom is 0, we need to use the F-distribution table with the largest possible degrees of freedom for the denominator, which is infinity.
Looking up the critical F-value for a right-tailed test and a significance level of 0.01 (α = 0.01) with numerator degrees of freedom = 40 and denominator degrees of freedom = infinity, we find that Fcrit ≈ 2.449.
Finally, we compare our calculated F-value (F = 1.782) with the critical F-value (Fcrit ≈ 2.449).
Since our calculated F-value is less than the critical F-value, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the variability in birth weight for preterm babies is significantly higher than that of full-term babies at the 0.01 level of significance.
In summary, based on the given information, we cannot support the obstetrician's claim that preterm babies have a higher variability of birth weight than full-term babies.