Testing Claims About Variation
In Exercises 5–16, test t...

Question

Testing Claims About Variation

In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Birth Weights A simple random sample of birth weights of 30 girls has a standard deviation of 829.5 g. Use a 0.01 significance level to test the claim that birth weights of girls have the same standard deviation as birth weights of boys, which is 660.2 g (based on Data Set 6 “Births” in Appendix B).

Accepted Answer

All right, hello, everyone. So this question says, a researcher collects a simple random sample of cholesterol levels from 25 adult men, and finds the sample standard deviation is 18.7 mg per deciliter. Test at the 0.01 significance level, the claim that the standard deviation of cholesterol levels in adult men is the same as the known standard deviation for adult women, which is 15.2 mg per deciliter. And here we have 4 different answer choices labeled A through D. All right, so first, what data do we actually have? Right? First we have N, which is our sample size of 25, and we also have S, which is the sample standard deviation. For men, which is equal to 18.7. Milligrams per deciliter. Now, we also have the claimed population standard deviation for women. So that's going to be labeled here as sigma. Or sigma not which was 15.2. Milligrams per deciliter. Now, the significance level, alpha is equal to 0.01. And in this case, the claim is that the standard deviations are equal, which means that we're using a two-tailed test. So let's test the variants, right? So for the null hypothesis. The sample variants Would correspond to. Sigma not squared, which is equal to 15.2 squared, and that should be equal to. 231.04. Now, the alternative hypothesis, on the other hand, states that sigma squared for men is not equal to 231.04. So now we need our chi squared test statistic. So our chi squared Is equal to And subtracted by 1 multiplied by s squared. Divided by Alpha knot squared, or excuse me, sigma not squared. So, plugging in the information we have already, that would be chi squared equals 25 subtracted by 1, multiplied by. 18.7 squared. And divided by 231.04. Evaluating this expression gives you a chi squared value equal to 36.32. So now it's 36.32 that we have to compare to our critical value. So for the critical value, we first have to find the degrees of freedom. And the degrees of freedom, if you recall, are equal to the sample size subtracted by 1, which in this case is 24. So, From the chi square distribution table, at alpha equals 0.01 for a two-tailed test, we get the following critical values. So the left critical value, which is. Chi squared at 0 point. 005 and 24 degrees of freedom. Is equal to approximately 10.856. Now for the right critical value, chi squared at 0 point. 995 and 24 degrees of freedom. Equals approximately 42.980. This leads you now to your decision. So now what is the rejection region? Well, the rejection region. Is So long as chi squared. Is less than the left critical value and or, or I should say, greater than the right critical value. So, because our test statistic of 36.32 is actually in between both critical values and not outside of the range. The test statistic does not fall in the rejection region, which means that we failed to reject the null hypothesis. So going back to our given answer choices. The correct one would be option B. Because we failed to reject the null hypothesis, there is not enough evidence to conclude a difference in standard deviations. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

Key Concepts

Hypothesis Testing

P-value

Standard Deviation and Variance

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