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Use a computer or calculator to complete the calculations and the hypothesis test for this exercise. Delco Products, a division of General Motors, produces commutators designed to be \(18.810 \mathrm{mm}\) in overall length. (A commutator is a device used in the electrical system of an automobile.) The following data are the lengths of a sample of 35 commutators taken while monitoring the manufacturing process: $$\begin{array}{lllllll}\hline 18.802 & 18.810 & 18.780 & 18.757 & 18.824 & 18.827 & 18.825 \\\18.809 & 18.794 & 18.787 & 18.844 & 18.824 & 18.829 & 18.817 \\\18.785 & 18.747 & 18.802 & 18.826 & 18.810 & 18.802 & 18.780 \\ 18.830 & 18.874 & 18.836 & 18.758 & 18.813 & 18.844 & 18.861 \\\18.824 & 18.835 & 18.794 & 18.853 & 18.823 & 18.863 & 18.808 \\\\\hline\end{array}$$ Is there sufficient evidence to reject the claim that these parts meet the design requirement "mean length is \(18.810 "\) at the \(\alpha=0.01\) level of significance?

Short Answer

Expert verified
Based on the calculated test statistic and the critical values, one can conclude whether or not enough evidence exists to reject the null hypothesis that the mean length of the commutators is \(18.810\) mm.

Step by step solution

01

Computing Sample Statistics

First, calculate the sample mean and the sample standard deviation of the 35 measurements. The sample mean \(\bar{x}\) is obtained by summing the lengths and dividing by the sample size. The sample standard deviation \(s\) is obtained from the formula \[s = \sqrt{\frac{1}{N-1} \sum_{i=1}^{N}(x_i-\bar{x})^2}\] where \(x_i\) are the lengths of the samples.
02

Setting up the Hypothesis

Set up the null and alternative hypotheses. The null hypothesis \(H_0\) is the claim that the mean length is 18.810 mm and the alternative hypothesis \(H_a\) is the claim that the mean length is not 18.810 mm. So, we have: \[H_0 : \mu = 18.810\] \[H_a : \mu \neq 18.810\]
03

Computing Test Statistics

Calculate the test statistic or z-score using the formula: \[z = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}\] where \(\mu_0\) is the value assumed in the null hypothesis, \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
04

Rejection Region and Conclusion

Determine the rejection region. Since the level of significance is \(\alpha = 0.01\), and the hypothesis is a two-tailed test, the rejection region is \(z < -2.58\) or \(z > 2.58\). Compare the calculated test statistic with the critical z-value. If the computed z-score falls in the rejection region then reject the null hypothesis. If not, fail to reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean Calculation
Understanding the sample mean is crucial when analyzing data sets, as it represents the average of the values. To calculate the sample mean, also known as the arithmetic mean, one must sum all the individual measurements of the sample and then divide the result by the total number of observations in the sample. In mathematical terms, if you have a sample with values denoted as \(x_1, x_2, \text{...,} x_n\), the sample mean, represented by \(\bar{x}\), is calculated using the formula: \[\bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i\] where \(n\) is the sample size, and \(\sum_{i=1}^{n}x_i\) is the sum of all sample values. In the given exercise involving commutators, the sample mean is calculated by adding together the lengths of the 35 commutators and then dividing by 35. This process gives an understanding of the typical length of a commutator produced by Delco Products.

Calculating the sample mean is the first step in various statistical analyses, including hypothesis testing. It allows us to establish a baseline from which we can determine if the observed sample differs significantly from the expected value.
Sample Standard Deviation
The sample standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

To calculate the sample standard deviation \(s\), you need to follow several steps. First, calculate the sample mean \(\bar{x}\) as described above. Next, subtract the sample mean from each individual value to get the deviations from the mean, and then square these deviations. Sum up all the squared deviations and divide this sum by \(n - 1\), which represents degrees of freedom in the sample. Finally, take the square root of the result. The formula is: \[s = \sqrt{\frac{1}{N-1} \sum_{i=1}^{N}(x_i-\bar{x})^2}\] In our exercise with commutators, \(N\) is 35. The standard deviation provides important context for the sample mean, offering insight into the reliability of the mean as a representation of the entire sample.
Z-Score Calculation
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. If a z-score is 0, it indicates that the data point's score is identical to the mean score. A z-score of 1.0 would indicate a value that is one standard deviation from the mean.

To calculate the z-score in hypothesis testing, use the formula: \[z = \frac{\bar{x}-\mu_0}{s/\sqrt{n}}\] where \(\mu_0\) is the mean value according to the null hypothesis (in our case, 18.810 mm), \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. The resulting z-score will tell you how many standard deviations the sample mean is from the hypothesized population mean.

In hypothesis testing, the z-score helps in deciding whether to reject the null hypothesis. It provides a standardized way of comparing the observed result with the expected result. If the z-score falls into the region beyond the predetermined threshold (the critical value), it suggests that the observed data is unlikely under the null hypothesis, leading to a rejection of the null hypothesis in the hypothesis test.

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Most popular questions from this chapter

Julia Jackson operates a franchised restaurant that specializes in soft ice cream cones and sundaes. Recently she received a letter from corporate headquarters warning her that her shop is in danger of losing its franchise because the average sales per customer have dropped "substantially below the average for the rest of the corporation." The statement may be true, but Julia is convinced that such a statement is completely invalid to justify threatening a closing. The variation in sales at her restaurant is bound to be larger than most, primarily because she serves more children, elderly, and single adults rather than large families who run up big bills at the other restaurants. Therefore, her average ticket is likely to be smaller and exhibit greater variability. To prove her point, Julia obtained the sales records from the whole company and found that the standard deviation was \(2.45\)dollar per sales ticket. She then conducted a study of the last 71 sales tickets at her store and found a standard deviation of \(2.95\)dollar per ticket. Is the variability in sales at Julia's franchise, at the 0.05 level of significance, greater than the variability for the company?

Calculate the \(p\) -value for each of the following hypothesis tests. a. \(H_{a}: \sigma^{2} \neq 20, n=15, \chi^{2} \star=27.8\) b. \(H_{a}: \sigma^{2}>30, n=18, \chi^{2} \star=33.4\) c. \(H_{a}: \sigma^{2} \neq 42,\) df \(=25, \chi^{2} \star=37.9\) d. \(H_{a}: \sigma^{2}<12,\) df \(=40, \chi^{2} \star=26.3\)

To test the hypothesis that the standard deviation on a standard test is \(12,\) a sample of 40 randomly selected students' exams was tested. The sample variance was found to be \(155 .\) Does this sample provide sufficient evidence to show that the standard deviation differs from 12 at the 0.05 level of significance?

In a sample of 60 randomly selected students, only 22 favored the amount budgeted for next year's intramural and interscholastic sports. Construct the \(99 \%\) confidence interval for the proportion of all students who support the proposed budget amount.

Acetaminophen is an active ingredient found in more than 600 over-the-counter and prescription medicines, such as pain relievers, cough suppressants, and cold medications. It is safe and effective when used correctly, but taking too much can lead to liver damage. A researcher believes the mean amount of acetaminophen per tablet in a particular brand of cold tablets is different from the 600 mg claimed by the manufacturer. A random sample of 30 tablets had a mean acetaminophen content of \(596.3 \mathrm{mg}\) with a standard deviation of \(4.7 \mathrm{mg}\). a. Is the assumption of normality reasonable? Explain. b. Construct a \(99 \%\) confidence interval for the estimate of the mean acetaminophen content. c. What does the confidence interval found in part b suggest about the mean acetaminophen content of one pill? Do you believe there is 600 mg per tablet? Explain.

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