/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The article "The Influence of Co... [FREE SOLUTION] | 91Ó°ÊÓ

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The article "The Influence of Corrosion Inhibitor and Surface Abrasion on the Failure of Aluminum-Wired Twist-On Connections" (IEEE Trans. on Components, Hybrids, and Manuf. Tech., 1984: 20-25) reported data on potential drop measurements for one sample of connectors wired with alloy aluminum and another sample wired with EC aluminum. Does the accompanying SAS output suggest that the true average potential drop for alloy connections (type 1) is higher than that for \(\mathrm{EC}\) connections (as stated in the article)? Carry out the appropriate test using a significance level of \(.01\). In reaching your conclusion, what type of error might you have committed? [Note: SAS reports the \(P\)-value for a two-tailed test.] \(\begin{array}{lrrrr}\text { Type } & \text { N } & \text { Mean } & \text { Std Dev } & \text { Std Error } \\ 1 & 20 & 17.49900000 & 0.55012821 & 0.12301241 \\ 2 & 20 & 16.90000000 & 0.48998389 & 0.10956373 \\ & & & & \\\ & & T & \text { DF } & \text { Prob> }|T| \\ \text { Variances } & & \text { T } & 0.0008 \\ \text { Unequal } & 3.6362 & 37.5 & 0.0008 .0008 \\ \text { Equal } & 3.6362 & 38.0 & 0.0000\end{array}\)

Short Answer

Expert verified
Yes, the test suggests the true average potential drop for alloy connections is higher. A Type I error might have been committed.

Step by step solution

01

State the Null and Alternative Hypotheses

We need to determine if the average potential drop for alloy connections is higher than for EC connections. The null hypothesis (\(H_0\)) will state that the true average potential drop for alloy connections is equal to or less than that for EC connections, i.e., \(\mu_1 \leq \mu_2\). The alternative hypothesis (\(H_a\)) will state that the average for alloy connections is higher, i.e., \(\mu_1 > \mu_2\).
02

Determine the Type of Test and Use One-Tailed Significance Level

The study compares means from two independent samples with unequal variances, evidenced by the output for 'variances = unequal.' The output provides a two-tailed \(P\)-value of 0.0008. Since we are interested in testing \(\mu_1 > \mu_2\), which is a one-tailed test, we will use half of that \(P\)-value: 0.0004. Our significance level is 0.01.
03

Compare the Calculated \(P\)-Value with the Significance Level

Our one-tailed \(P\)-value is 0.0004, which is less than the significance level of 0.01. Thus, we reject the null hypothesis \(H_0\) in favor of the alternative hypothesis \(H_a\), suggesting that the true average potential drop for alloy connections is significantly higher than that for EC connections.
04

Consider Possible Errors

Since we rejected the null hypothesis, there is a potential for committing a Type I error. This type of error occurs when we incorrectly reject a true null hypothesis. In this context, it means concluding that the potential drop for alloy connections is higher when it actually isn't. The risk of this error is equal to our significance level, 0.01.

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Key Concepts

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

Type I Error
In the realm of hypothesis testing, a Type I error represents the mistake of rejecting a true null hypothesis. Imagine a situation where you incorrectly conclude that a treatment or condition has an effect when, in reality, it does not. This can be quite misleading, leading to incorrect claims or unnecessary actions.
  • A Type I error is like a false alarm, assuming something is happening when it isn't.
  • The probability of committing a Type I error is defined by the significance level, which in this exercise is 0.01.
  • In our context, if the null hypothesis asserted that the average potential drop for alloy connections is not higher, rejecting this when it's true would be a Type I error.
Thus, always consider the risk of a Type I error when interpreting results, especially since this risk is expressed by the significance level chosen for the test.
P-Value
The p-value in hypothesis testing quantifies the evidence against the null hypothesis. It's a probability that measures how well your data align with the null hypothesis. A lower p-value indicates stronger evidence against the null hypothesis.
  • In simple terms, the p-value asks, "How rare is this data, assuming the null hypothesis is true?"
  • During the exercise, we derived a one-tailed p-value of 0.0004, which is quite small, suggesting that our observed data would be very unusual if the null hypothesis were true.
  • This led us to reject the null hypothesis in favor of the alternative hypothesis because the p-value was smaller than our significance level of 0.01.
Therefore, p-values are vital in showing whether our results are statistically significant.
Significance Level
The significance level, often denoted by alpha (\(\alpha\)), is pre-determined before conducting a test and represents the tolerance for Type I error in decision-making. It is the threshold probability for rejecting the null hypothesis.
  • In our example, a significance level of 0.01 means there's a 1% risk of mistakenly rejecting the true null hypothesis.
  • Choosing the significance level reflects the degree of confidence you require for your results. Lower levels mean stricter criteria for rejecting the null hypothesis.
  • When the p-value is less than the significance level, it supports rejecting the null hypothesis.
Utilized correctly, significance levels help balance the risk of Type I errors while assessing evidence against the null hypothesis.
Null Hypothesis
The null hypothesis, denoted as (\(H_0\)), is the default assumption or starting claim of a hypothesis test. It suggests that there is no effect or difference, serving as a basis for statistical testing.
  • For the exercise in question, the null hypothesis posited that the true average potential drop is less than or equal for alloy compared to EC connections (\(\mu_1 \leq \mu_2\)).
  • The job of hypothesis testing is to evaluate whether the data provides strong enough evidence to reject this assumption.
  • The null hypothesis is not proven true through testing; it can either be rejected or not rejected based on compared statistics.
Understanding the null hypothesis as a central pillar in statistical analysis assists in assessing the differences in the data.
Alternative Hypothesis
An alternative hypothesis (\(H_a\)) is what statisticians consider when evidence suggests the null hypothesis may not be true. It proposes that there is an effect or difference, essentially the opposite of the null hypothesis.
  • In the provided exercise, the alternative hypothesis declared that the average potential drop for alloy connections is indeed greater than that for EC connections (\(\mu_1 > \mu_2\)).
  • Evidence against the null hypothesis is interpreted as support for the alternative hypothesis.
  • When the p-value is sufficiently low to reject the null hypothesis, it lends credence to the alternative hypothesis.
By structuring hypotheses correctly, we ensure a logical framework for assessing and understanding experimental data.

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Most popular questions from this chapter

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar{x}=18.12 \mathrm{kgf} / \mathrm{cm}^{2}\) for the modified mortar ( \(m=40\) ) and \(\bar{y}=16.87 \mathrm{kgf} / \mathrm{cm}^{2}\) for the unmodified mortar ( \(n=32\) ). Let \(\mu_{1}\) and \(\mu_{2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. a. Assuming that \(\sigma_{1}=1.6\) and \(\sigma_{2}=1.4\), test \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{2}: \mu_{1}-\mu_{2}>0\) at level .01. b. Compute the probability of a type II error for the test of part (a) when \(\mu_{1}-\mu_{2}=1\). c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta=, 10\) when \(\mu_{1}-\mu_{2}=1\). If \(m=40\), what value of \(n\) is necessary? d. How would the analysis and conclusion of part (a) change if \(\sigma_{1}\) and \(\sigma_{2}\) were unknown but \(s_{1}=1.6\) and \(s_{2}=1.4\) ?

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). The article "A Gap Detection Tactility Test for Sensory Deficits Associated with Carpal Tunnel Syndrome" (Ergonomics, 1995: 2588-2601) reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for \(m=8\) normal subjects was \(1.71 \mathrm{~mm}\), and the sample standard deviation was 53 ; for \(n=10\) CTS subjects, the sample mean and sample standard deviation were \(2.53\) and \(.87\), respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of \(.01\).

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: After \(S \quad F\) where \(x_{1}+x_{2}+x_{3}+x_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\) denote the four cell probabilities, so that \(p_{1}=P(S\) before and \(S\) after), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{f}-X_{j}\right) / n\) has approximately a normal distribution with variance given by \(\left[p_{i}+p_{j}-\left(p_{j}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called McNemar's test). d. If \(x_{1}=350, \quad x_{2}=150, \quad x_{3}=200\), and \(x_{4}=300\), what do you conclude?

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution. \begin{tabular}{lrr} & \(\mathbf{A}\) & \(\mathbf{B}\) \\ \hline \(26-<30\) & 6 & 4 \\ \(30-<34\) & 12 & 9 \\ \(34-<38\) & 15 & 19 \\ \(38-<42\) & 7 & 10 \\ & \(m=40\) & \(m=42\) \\ \hline \end{tabular} Compute a \(95 \%\) CI for the difference between the true proportions of all specimens of alloys \(\mathrm{A}\) and \(\mathrm{B}\) that have an ultimate strength of at least \(34 \mathrm{ksi}\).

An experiment to determine the effects of temperature on the survival of insect eggs was described in the article "Development Rates and a Temperature- Dependent Model of Pales Weevil" (Environ. Entomology, 1987:956-962). At \(11^{\circ} \mathrm{C}, 73\) of 91 eggs survived to the next stage of development. At \(30^{\circ} \mathrm{C}, 102\) of 110 eggs survived. Do the results of this experiment suggest that the survival rate (proportion surviving) differs for the two temperatures? Calculate the \(P\)-value and use it to test the appropriate hypotheses.

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