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Scientists think that robots will play a crucial role in factories in the next several decades. Suppose that in an experiment to determine whether the use of robots to weave computer cables is feasible, a robot was used to assemble 500 cables. The cables were examined and there were 15 defectives. If human assemblers have a defect rate of \(.035\) \((3.5 \%)\), does this data support the hypothesis that the proportion of defectives is lower for robots than for humans? Use a .01 significance level.

Short Answer

Expert verified
The data does not support the hypothesis that robots have a lower defect rate than humans.

Step by step solution

01

Define Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis (H_0) states that the defect rate for robots, \( p \), is greater than or equal to the human defect rate, i.e., \( p \geq 0.035 \). The alternative hypothesis (H_a) states that the defect rate for robots is lower than that for humans, i.e., \( p < 0.035 \).
02

Calculate Sample Proportion

To find the sample proportion of defective cables made by the robot, divide the number of defectives by the total number of cables: \( \hat{p} = \frac{15}{500} = 0.03 \).
03

Determine the Standard Error

The standard error of the sample proportion is calculated by \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( n \) is the sample size and \( p \) is the hypothesized proportion of 0.035. Here, \( SE = \sqrt{\frac{0.035(1-0.035)}{500}} \approx 0.0085 \).
04

Calculate the Z-Score

The Z-score is found using the formula: \( Z = \frac{\hat{p} - p}{SE} \). Substitute \( \hat{p} = 0.03 \), \( p = 0.035 \), and \( SE = 0.0085 \):\( Z = \frac{0.03 - 0.035}{0.0085} \approx -0.588 \).
05

Decide Based on Significance Level

With a significance level \( \alpha = 0.01 \), we find the critical Z-value for a one-tailed test, which is \( -2.33 \). Since \( -0.588 \) is greater than \( -2.33 \), we 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.

Significance Level
In hypothesis testing, the significance level is a crucial concept. It helps us decide how strong or convincing our data needs to be to support our conclusion. The significance level, often denoted by \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it is the risk we are willing to take to make a mistake in our inference.

A common significance level is \( 0.01 \), which implies we are willing to accept a 1% chance of a Type I error, or a false positive. In our case, scientists use a significance level of \( 0.01 \) to determine if the robot-made cables are less defective than human-made ones. This means scientists are quite stringent, requiring strong evidence to claim robots are indeed better.
  • A lower \( \alpha \) value indicates more stringent criteria for accepting an alternative hypothesis.
  • If our test statistic, such as a Z-score, falls beyond the critical value corresponding to \( \alpha \), it supports the alternative hypothesis.
  • The \( \alpha \) value affects the test's power and error rates, balancing the risks of wrong conclusions.
Alternative Hypothesis
The alternative hypothesis is a statement that proposes a difference, effect, or relationship. It is what we aim to support through our data analysis. In hypothesis testing, we compare this against the null hypothesis, which is usually a statement of no effect or difference.

In our scenario, the alternative hypothesis (\( H_a \)) is that robots make cables with a lower defect rate than humans (\( p < 0.035 \)). This statement suggests a potential improvement with robot assembly, which researchers want to prove as true. The alternative hypothesis stands in contrast to the null hypothesis, which suggests that robot defects are the same or worse.
  • The alternative hypothesis can be one-sided (like ours) or two-sided, depending on the nature of the statement.
  • It influences how we calculate and interpret test statistics like the Z-score.
  • Acceptance of the alternative hypothesis implies a statistical significance in findings.
Standard Error
Standard error is an essential component in statistics, representing the average distance between a sample statistic and the population parameter it estimates. Specifically, it measures how much the sample proportion is expected to fluctuate around the true population proportion.

In our example, the standard error helps us understand the variability in the defect rates from the robot-assembled cables. It is calculated using the formula:\[ SE = \sqrt{\frac{p(1-p)}{n}} \]. Here, \( p \) is the defect rate (0.035 for humans), and \( n \) is the number of cables (500). For our test, \( SE \approx 0.0085 \).
  • Smaller standard errors suggest more precise estimates of the population parameter.
  • As sample size \( n \) increases, the standard error generally decreases, suggesting less variation.
  • Standard error impacts the width of confidence intervals and the calculation of test statistics like Z-scores.
Z-score
The Z-score is an essential tool in hypothesis testing to determine how many standard deviations an element is from the mean. In our analysis, it helps compare the sample proportion to the hypothesized population proportion.

To calculate the Z-score, we apply the formula: \( Z = \frac{\hat{p} - p}{SE} \), where \( \hat{p} \) is the observed sample proportion (0.03), \( p \) is the hypothesized proportion (0.035), and \( SE \) is the standard error (0.0085). In this case, the Z-score is approximately \(-0.588\).
  • A Z-score helps identify whether the observed result is statistically significant.
  • If the Z-score falls into the critical region determined by the significance level, we reject the null hypothesis.
  • In our example, since \(-0.588\) is greater than the critical Z-value of \(-2.33\), we cannot conclude the defect rate for robots is indeed lower.

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Most popular questions from this chapter

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various consumer products. Suppose five \(6.0 \mathrm{oz}\) tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data (consistent with what the cited article reported): \(.53, .65, .46, .50, .37\). Does it appear that the true average amount left is less than \(10 \%\) of the advertised net contents? a. Check the validity of any assumptions necessary for testing the appropriate hypotheses. b. Carry out a test of the appropriate hypotheses using a significance level of \(.05\). Would your conclusion change if a significance level of .01 had been used? c. Describe in context type I and II errors, and say which error might have been made in reaching a conclusion.

Chapter 7 presented a CI for the variance \(\sigma^{2}\) of a normal population distribution. The key result there was that the rv \(\chi^{2}=(n-1) S^{2} / \sigma^{2}\) has a chi-squared distribution with \(n-1\) df. Consider the null hypothesis \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) (equivalently, \(\left.\sigma=\sigma_{0}\right)\). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) S^{2} / \sigma_{0}^{2}\) has a chi-squared distribution with \(n-1\) df. If the relevant alternative is \(H_{\mathrm{a}}: \sigma^{2}>\sigma_{0}^{2}\), rejecting \(H_{0}\) if \((n-1) s^{2} / \sigma_{0}^{2} \geq \chi_{\alpha, n-1}^{2}\) gives a test with significance level \(\alpha\). To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most \(.50^{\circ} \mathrm{C}\). The softening points of ten different specimens were determined, yielding a sample standard deviation of \(.58^{\circ} \mathrm{C}\). Does this strongly contradict the uniformity specification? Test the appropriate hypotheses using \(\alpha=.01\).

A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood. Does this suggest that the actual percentage of type A donations differs from \(40 \%\), the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of \(.01\). Would your conclusion have been different if a significance level of \(.05\) had been used?

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at \(40 \mathrm{mph}\) under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following three rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, \quad R_{2}=\\{\bar{x}: \bar{x} \leq 115.20\\}\), \(R_{3}=\\{\bar{x}\) : either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\}\) ? c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z: z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\}\) ?

Exercise 36 in Chapter 1 gave \(n=26\) observations on escape time \((\mathrm{sec})\) for oil workers in a simulated exercise, from which the sample mean and sample standard deviation are \(370.69\) and \(24.36\), respectively. Suppose the investigators had believed a priori that true average escape time would be at most \(6 \mathrm{~min}\). Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using a significance level of .05.

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