Tennis Ball Manufacturing A company manufactures tennis balls....

Question

Tennis Ball Manufacturing A company manufactures tennis balls. When the balls are dropped onto a concrete surface from a height of 100 inches, the company wants the mean bounce height to be 55.5 inches. This average is maintained by periodically testing random samples of 25 tennis balls. If the t-value falls between and , then the company will be satisfied that it is manufacturing acceptable tennis balls. For a random sample, the mean bounce height of the sample is 56.0 inches and the standard deviation is 0.25 inch. Assume the bounce heights are approximately normally distributed. Is the company making acceptable tennis balls? Explain.

Accepted Answer

Hi, everyone. Glad to have you back. Our next problem says A snack company produces potato chips. The company claims that the mean weight of a bag is 50 g. To monitor quality, a random sample of 16 bags is selected, and the mean weight is found to be 51 g, with a standard deviation of 0.4 g. Assume the weights are approximately normally distributed. Using a significance level of alpha equals 0.05, two-tailed, are the bags meeting the claimed mean weight. And then in parentheses critical T values or degrees of freedom equaling 15, is approximately plus or minus 2.131. And we have answer choices A, yes, the bags meet the claimed mean weight, B, no, the bags do not meet the claimed mean weight, C, not enough information to decide, or D, the sample size is too small to test. Well, we can eliminate choice D right away. We are able to use this number of samples, especially because we're told that the weights are normally distributed. So, let's figure out, it certainly seems as if we have enough information to decide. We are given our sample number, our population mean. And our sample mean and standard deviation, so we should have plenty of information. To decide, so cross out choice C. So we just need to see if the claim is or is not correct. So first off, we need to think about our null hypothesis and alternative hypothesis. So, our company is claiming that the mean weight is 50 g, so we would say that the claim Is. That mu equals 50. Now, this is a statement of equality, and so this will be our null hypothesis in this case. So even though that's the claim, we treat it as the null hypothesis because this has the statement of equality. And that means our alternative hypothesis would be that me is not equal to 50. So in this case we're talking about a two-tailed. Test Now we have fewer than 20. Values in our sample, but we are told we can assume a normal distribution. So we can use our test statistic based on that. So, in that case, we are looking for the mean. But the Population standard deviation, so sigma is not known. So in that case, the test statistic we are looking for is T. So we can compute that test statistic. So T is equal to, and we have the formula. X bar Minus mu in the numerator and then in the denominator. S divided by the square root of N. So we have all this information, so indeed there's enough information. So, let's plug in our numbers that we have. So, X bar or sample mean, was 51 g. So, in the numerator, 51 minus the population mean mu, which you said is equal to 50. That's the numerator, and then the denominator, we have the uh standard deviation of our sample. So, S, uh we're told that that's 0.4. And we're dividing that by the square root of N, the sample number. Which we are told there was a sample of 16 bags, so, square root of 16. Now we can actually simplify that expression quite easily to one. Divided by 0.1, which is just going to equal 10. Now, this problem was made a little easier for us by the fact that we're given the critical T values that we compare this to. So, since we have 16 bags in our sample, we know that our degrees of freedom is 15, and we're told that our critical T values for 15, are approximately plus or minus 2.131. So we look at whether our T value is below the negative critical T value or above the positive critical T value. So as we said, our T-value is 10, which is greater than 2.131. Which is that positive critical T value. And so as a result, we reject. The null hypothesis. Since the null hypothesis. Was that the mean is 50 then we. Reject the claim that that is the case. So, no, the bags do not meet the claimed weight, and our answer is choice B. Cross out choice A. So, once again, given the information we have here, our company's claiming the mean weight is 50 g, we have a sample of 16 with the mean weight of 51 g, and we're given a standard deviation. We're told to assume normal distribution, we have our significance level, and we are asked to determine if the bags are meeting the claimed mean weight. And we calculated our T value because we were told to assume normal distribution. And that T value was above the critical T value, greater than the critical T value. Therefore, we reject our null hypothesis, and we can conclude that choice B, no, the bags do not meet the claimed mean weight. See you in the next video.

Key Concepts

Sampling Distribution

Hypothesis Testing

Confidence Intervals

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