A nutrition bar manufacturer claims that the standard deviatio...

Question

A nutrition bar manufacturer claims that the standard deviation of the number of grams of carbohydrates in a bar is 1.11 grams. A random sample of 26 bars has a standard deviation of 1.19 grams. At α=0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed.

Accepted Answer

All right, hi everyone. So this question says, a furniture maker claims that the standard deviation of oak plank thickness is 0.05 centimeters. A random sample of 20 planks yields a sample standard deviation of 0.07 centimeters. Assume thickness is normally distributed. At alpha equals 0.05, is there sufficient evidence to reject the maker's claim? And here we have 4 different answer choices labeled A through D. So, first and foremost, what are the hypotheses that we are? Working with here. Well, notice the wording of the question. The question is asking us if we can reject the claim that the maker is making. Because of that, the claim should be the null hypothesis. So each knot. would state that sigma, the standard deviation, is 0.05. This means that H A, the alternative, would state the opposite, so that sigma is not equal to 0.05. So now let's move on to our test statistic. Now our chi square test statistic is equal to and subtracted by 1. Multiplied by squared. Divided by Sigma not swear. So N is the sample size, S is the sample standard deviation, and sigma squared, excuse me, sigma knot is the claimed population standard deviation. So plugging in the information that you have, that is. 20 subtracted by one. Multiplied by 0.07 squared. And divided by. 0.05 squared. Evaluating this expression gives you a chi square test statistic equal to 37.24. And now you can find the critical values to compare against the test statistic. So, first, recall that the degrees of freedom is equal to the sample size and subtracted by 1, which is equal to 19. So the chi square critical value. At alpha divided by 2, which is 0.025 and 19 degrees of freedom. Is approximately equal to. 8.907. But recall that this is a two-tailed test, so you have to account for the other tail as well. So for the other tale, the critical value. For chi square at 0.975 and 19 degrees of freedom. is equal to 32.852. And now we're left with our decision. Notice how our test statistic, 37.24 is greater than. Our larger critical value, 32.852. Because of this, It is within the rejection region, which means that we can reject the null hypothesis. So what does this tell you? What conclusion can you now make? At this 5% significance level, there is enough evidence to conclude that the true standard deviation differs from 0.05 centimeters. This means that option A is our correct answer. 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

Standard Deviation

Hypothesis Testing

Significance Level (α)

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