Technology
In Exercises 9–12, test the given claim by us...

Question

Technology

In Exercises 9–12, test the given claim by using the display provided from technology. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Peanut Butter Cups Data Set 38 “Candies” includes weights of Reese’s peanut butter cups. The accompanying Statdisk display results from using all 38 weights to test the claim that the sample is from a population with a mean equal to 8.953 g.

Accepted Answer

Welcome back, everyone. In this problem, a simple random sample of 45 chocolate chip cookies from a bakery is analyzed by a nutritionist for sugar content in grams. The bakery claims that the average sugar content is 12.5 g per cookie. At a 0.05 significance level, the following analysis output was obtained from a statistical software. The test statistic T is -3.46821. The criticality value is plus R minus 2.01537 and the P value is 0.00118. Using the output, what should the nutritionist conclude about the bakery's claim that the mean sugar content is 12.5 g? Now in this problem we are testing a claim of about a population mean and where we don't know the population standard deviation. The only thing we know so far is that the sample is a simple random sample and the sample size of 45 is greater than 30. So let's go ahead and first start by stating our hypothesis. Remember, we are testing the claim that the mean sugar content is 12.5 g. That means or another hypothesis is that the mean sugar content mu is 12 equals 12.5, OK. Which then means that the alternative hypothesis is that the mean sugar content is not 12.5 g, so that mal is not equal to 12.5. Now, if we think about it here, since we're testing if it is a value or not, then this is a two-tailed test. And already in this T test, we're told that the test statistic is -3.46821 and the critical T value is plus R minus 2.01537. So if we were to put that on a distribution here, OK, then for our critical T value in our two-tail test. We are first testing whether the test statistic is inside or, sorry, it's less than or greater than the absolute value of 2.01537. So we're trying to figure out if it's greater than 2.01537 or less than negative 2.01537, OK? And what we're saying based on these hypotheses is that if it is Outside of these values, or if the p value is less than alpha, we reject our null hypothesis. So in these areas that are shaded red here, if our test statistic is outside these values, we would reject, OK? And if it's inside, then we don't have enough evidence to reject the null hypothesis. Now remember we're told the test statistic is -3.468. So on our distribution, it would fall here. This would be our T value. And no, because of that T value. OK, then we know that we are going to reject the null hypothesis, OK? In other words, since T, OK. Which is, or rather, let's work with the absolute value here. So the absolute value of T, OK, which would be 3.46821, is greater than the absolute value of our critical value, 2.01537. OK. And All right, well, this, based on this condition actually, we can reject the null hypothesis, OK? We reject the null hypothesis. Now remember, there's another condition we can test. We can also use what we know about the p value. Remember, That here for this significance level, if our p value is less than the significant level alpha, we can also reject the non hypothesis. So, here, notice that our P value. Which is 0.0018118. OK, is less than our alpha or value of alpha, which is 0.05. So, this also further strengthens our decision or supports our decision to reject the null hypothesis. Therefore, what can we say? Well, let's write out our statement because at the 0.05 significance level. There is sufficient evidence. Sufficient evidence. To warrant a rejection, to warrant a rejection of the bakery's claim. Of the bakery's claim that the mean sugar content. Is 12.5 g. Uh cookie, yeah. So that means the nutritionist should conclude that the average sugar content is significantly different from 12.5 g. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Null and Alternative Hypotheses

Test Statistic

P-Value

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