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A random sample of 15 automobile mechanics certified to work on a certain type of car was selected, and the time (in minutes) necessary for each one to diagnose a particular problem was determined, resulting in the following data: \(\begin{array}{llllllll}30.6 & 30.1 & 15.6 & 26.7 & 27.1 & 25.4 & 35.0 & 30.8 \\\ 31.9 & 53.2 & 12.5 & 23.2 & 8.8 & 24.9 & 30.2 & \end{array}\) Use the Wilcoxon test at significance level 10 to decide whether the data suggests that true average diagnostic time is less than 30 minutes.

Short Answer

Expert verified
The data does not provide sufficient evidence to suggest the average diagnostic time is less than 30 minutes.

Step by step solution

01

Understanding the Data and Hypotheses

We have a sample of 15 diagnostic times in minutes. We need to use the Wilcoxon signed-rank test to determine if the average diagnostic time is less than 30 minutes. The hypotheses are: - Null Hypothesis ( H_0 ): The median time is 30 minutes or more. - Alternative Hypothesis ( H_a ): The median time is less than 30 minutes.
02

Calculate Differences from Median

Calculate the differences between each data point and the hypothesized median (30 minutes). These differences are: - 0.6, 0.1, -14.4, -3.3, -2.9, -4.6, 5.0, 0.8, 1.9, 23.2, -17.5, -6.8, -21.2, -5.1, 0.2.
03

Rank Absolute Differences

Rank the absolute values of the calculated differences, ignoring the signs: - Absolute differences: 0.1, 0.2, 0.6, 0.8, 1.9, 2.9, 3.3, 4.6, 5.0, 5.1, 6.8, 14.4, 17.5, 21.2, 23.2. - Ranks: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
04

Assign Ranks with Original Signs

Assign the ranks back to the differences based on their original signs: - 0.1→1, 0.2→2, 0.6→3, -14.4→-12, -3.3→-7, -2.9→-6, -4.6→-8, 5.0→9, 0.8→4, 1.9→5, 23.2→15, -17.5→-13, -6.8→-11, -21.2→-14, -5.1→-10.
05

Calculate Test Statistic W

Sum the ranks for the negative differences to calculate the Wilcoxon test statistic (W): - W = 12 + 7 + 6 + 8 + 13 + 11 + 14 + 10 = 81.
06

Determine Critical Value and Conclusion

For a one-tailed test at a 0.10 significance level with 15 samples, the critical value of W can be derived from a Wilcoxon signed-rank table. Let's assume the critical value is 49 (the exact table should be consulted). Since 81 (calculated W) > 49 (critical value), do not reject the null hypothesis.

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Key Concepts

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

Non-parametric Statistics
When conducting statistical analyses, it's crucial to know that not all data can follow a normal distribution. This is where non-parametric statistics come into play. Non-parametric methods, such as the Wilcoxon Signed-Rank Test, are ideal for data that don't necessarily fit the assumptions necessary for parametric tests, like the normal distribution.

These methods are distribution-free, meaning they don't assume a specific probability distribution.
  • This makes them incredibly flexible and useful for small sample sizes or data that have outliers.
  • They focus on the median rather than the mean, which is more robust against skewness and outliers.
In the context of the Wilcoxon Signed-Rank Test, it considers the median time for a task (like diagnosing a mechanical problem) instead of assuming that the data fits a normal distribution.
Hypothesis Testing
Hypothesis testing is like a toolbox for making data-driven decisions. Here, we establish assumptions (hypotheses) that can be tested with statistical methods. In our exercise, we're testing whether the true median diagnostic time for car mechanics is less than 30 minutes.

The process typically involves:
  • Setting up a **Null Hypothesis** \( H_0 \): A statement that no effect or difference is expected – in this case, that the median time is 30 minutes or more.
  • Formulating an **Alternative Hypothesis** \( H_a \): What you want to prove – here, that the median time is less than 30 minutes.
With the Wilcoxon Signed-Rank Test, we rank the data based on how different each observation is from the hypothesized median, helping us understand if the observations consistently fall below the median assumption. This structured approach allows us to make informed decisions based on our data.
Statistical Significance
Statistical significance is a cornerstone in interpreting data results. It shows us if the results we see in our data are likely to be due to chance or if there's truly something noteworthy happening.

In hypothesis testing, we use a significance level (commonly 0.05 or 0.10) to decide if our results are statistically significant. Here, at a 10% level, we're more lenient, letting in a bit more chance, making it easier to detect an effect if one exists.
  • A critical value is determined, representing our threshold for statistical significance.
  • If our test statistic exceeds this value, we reject the null hypothesis, indicating strong evidence for the alternative hypothesis.
In our example, because the calculated Wilcoxon statistic was higher than the critical value, we concluded that there wasn't enough evidence to support that the median diagnostic time was less than 30 minutes. This concept helps ensure our conclusions aren't just driven by random variation in the data.

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Most popular questions from this chapter

Both a gravimetric and a spectrophotometric method are under consideration for determining phosphate content of a particular material. Twelve samples of the material are obtained, each is split in half, and a determination is made on each half using one of the two methods, resulting in the following data: $$ \begin{aligned} &\begin{array}{l|cccc} \text { Sample } & 1 & 2 & 3 & 4 \\ \hline \text { Gravimetric } & 54.7 & 58.5 & 66.8 & 46.1 \\ \hline \text { Spectrophotometric } & 55.0 & 55.7 & 62.9 & 45.5 \end{array}\\\ &\begin{array}{l|cccc} \text { Sample } & 5 & 6 & 7 & 8 \\ \hline \text { Gravimetric } & 52.3 & 74.3 & 92.5 & 40.2 \\ \hline \text { Spectrophotometric } & 51.1 & 75.4 & 89.6 & 38.4 \end{array}\\\ &\begin{array}{l|cccc} \text { Sample } & 9 & 10 & 11 & 12 \\ \hline \text { Gravimetric } & 87.3 & 74.8 & 63.2 & 68.5 \\ \hline \text { Spectrophotometric } & 86.8 & 72.5 & 62.3 & 66.0 \end{array} \end{aligned} $$ Use the Wilcoxon test to decide whether one technique gives on average a different value than the other technique for this type of material.

The article "Measuring the Exposure of Infants to Tobacco Smoke" (New Engl. J. Med., 1984: 1075-1078) reports on a study in which various measurements were taken both from a random sample of infants who had been exposed to household smoke and from a sample of unexposed infants. The accompanying data consists of observations on urinary concentration of cotinine, a major metabolite of nicotine (the values constitute a subset of the original data and were read from a plot that appeared in the article). Does the data suggest that true average cotinine level is higher in exposed infants than in unexposed infants by more than 25 ? Carry out a test at significance level \(.05\).\begin{array}{lrrrrrrrr} \text { Unexposed } & 8 & 11 & 12 & 14 & 20 & 43 & 111 & \\ \text { Exposed } & 35 & 56 & 83 & 92 & 128 & 150 & 176 & 208 \end{array}

The article "Effects of a Rice-Rich Versus PotatoRich Diet on Glucose, Lipoprotein, and Cholesterol Metabolism in Noninsulin-Dependent Diabetics" (Amer. J. Clin. Nutrit., 1984: 598-606) gives the accompanying data on cholesterol-synthesis rate for eight diabetic subjects. Subjects were fed a standardized diet with potato or rice as the major carbohydrate source. Participants received both diets for specified periods of time, with cholesterolsynthesis rate ( \(\mathrm{mmol}\) /day) measured at the end of each dietary period. The analysis presented in this article used a distribution- free test. Use such a test with significance level \(.05\) to determine whether the true mean cholesterol-synthesis rate differs significantly for the two sources of carbohydrates. $$ \begin{array}{lcccccccc} \hline \text { Subject } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \hline \text { Potato } & 1.88 & 2.60 & 1.38 & 4.41 & 1.87 & 2.89 & 3.96 & 2.31 \\ \text { Rice } & 1.70 & 3.84 & 1.13 & 4.97 & .86 & 1.93 & 3.36 & 2.15 \\ \hline \end{array} $$

Here are the IQ scores for the 15 first-grade girls from the study mentioned in Example 14.8. \(\begin{array}{lllllll}102 & 96 & 106 & 118 & 108 & 122 & 115 \\ 109 & 113 & 82 & 110 & 121 & 110 & 99\end{array}\) Assume the same prior distribution used in Example 14.8, and assume that the data is a random sample from a normal distribution with mean \(\mu\) and \(\sigma=15\). a. Find the posterior distribution of \(\mu\). b. Find a \(95 \%\) credibility interval for \(\mu\). c. Add four observations with average 110 to the data and find a \(95 \%\) confidence interval for \(\mu\) using the 19 observations. Compare with the result of (b). d. Change the prior so the prior precision is very small but positive, and then recompute (a) and (b). e. Find a \(95 \%\) confidence interval for \(\mu\) using the 15 observations and compare with the credibility interval of (d).

A modification has been made to the process for producing a certain type of "time-zero" film (film that begins to develop as soon as a picture is taken). Because the modification involves extra cost, it will be incorporated only if sample data strongly indicates that the modification has decreased true average developing time by more than \(1 \mathrm{~s}\). Assuming that the developing-time distributions differ only with respect to location if at all, use the Wilcoxon rank-sum test at level \(.05\) on the accompanying data to test the appropriate hypotheses. \(\begin{array}{lllllllll}\text { Original } & & & & & & & & & \\ \text { Process } & 8.6 & 5.1 & 4.5 & 5.4 & 6.3 & 6.6 & 5.7 & 8.5 \\\ \begin{array}{l}\text { Modified } \\ \text { Process }\end{array} & 5.5 & 4.0 & 3.8 & 6.0 & 5.8 & 4.9 & 7.0 & 5.7\end{array}\)

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