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.

Minting of Pennies Data Set 40 “Coin Weights” lists weights (grams) of pennies minted after 1983. Here are the statistics for those weights: n = 37, xbar = 2.49910 g, s = 0.01648 g . Use a 0.05 significance level to test the claim that the sample is from a population of pennies with weights having a standard deviation greater than 0.01000 g.

Accepted Answer

Hello everyone. Let's take a look at this question together. A manufacturer produces bottles of olive oil labeled as containing 500 mL. A quality control specialist wants to ensure the consistency of the filling process. From a simple random sample of 30 bottles, the following statistics are obtained. N is equal to 30, X bar is equal to 498.7 mL, and S is equal to 2.05 mL. Assume the population of fill volumes is normally distributed. At the 0.05 significance level, test the claim that the standard deviation of the fill volume is less than 3 mL. Is it answer choice A, do not reject the null hypothesis, as there is not enough evidence to support the claim. Answer choice B, reject the null hypothesis as the standard deviation is less than 3 mL. Answer choice C, reject the null hypothesis, as there is enough evidence to support that the standard deviation is less than 3 mL or answer choice D do not. the null hypothesis as the standard deviation equals 3 mL. So in order to solve this question, we have to recall how we can test the claim that the standard deviation of the fill volume is less than 3 mL using the given data, which the given data includes a sample size of 30, a sample standard deviation of 2.05 mL. A claimed population standard deviation of 3 mL, a significance level of 0.05, as well as the population is normally distributed. So we are testing if the standard deviation is less than 3 mL, so we test the hypotheses, which include the null hypothesis of the standard deviation being equal to 3 mL. Or the alternate hypothesis of the standard deviation being less than 3 mL, and we know that this is a left tailed test. So then we use the chi square test statistic for variances using the formula for the chi square test statistic, and plugging in our values, we get 13.54 as the value for the chi square test. And then we can find the critical value, which we know that our degrees of freedom is equal to N minus 1, and since our value for N is 30, N minus 1 is equal to 29, so our degrees of freedom is 29, as well as we have that significance level of 0.05, and using a chi square table or calculator, we get Our chi square critical value of 17.708, which we can then compare the test statistic with the critical value for the chi square, which based on that comparison, we know that we are in the rejection region. And so we reject the null hypothesis. And since we reject the null hypothesis, we know that there is sufficient evidence at the 0.05 significant level to support the claim that the standard deviation of the fill volume is less than 3 mL, which makes answer choice C the correct answer. And all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Hypothesis Testing

P-value

Standard Deviation

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