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A semiconductor manufacturer collects data from a new tool and conducts a hypothesis test with the null hypothesis that a critical dimension mean width equals \(100 \mathrm{nm}\). The conclusion is to not reject the null hypothesis. Does this result provide strong evidence that the critical dimension mean equals \(100 \mathrm{nm}\) ? Explain.

Short Answer

Expert verified
Not rejecting the null hypothesis does not provide strong evidence that the mean equals 100 nm; it simply lacks evidence to refute this value.

Step by step solution

01

Understanding the Hypothesis Test

The hypothesis test involves a null hypothesis and an alternative hypothesis. The null hypothesis (H0) is that the critical dimension mean width equals 100 nm (\[ H_0: \mu = 100 \, \text{nm} \]). The alternative hypothesis (H1) would be that the mean does not equal 100 nm (\[ H_1: \mu eq 100 \, \text{nm} \]).
02

Analyzing the Conclusion

The conclusion is to not reject the null hypothesis. This means that based on the data and the level of significance used, there is not enough statistical evidence to conclude that the critical dimension mean width is different from 100 nm.
03

Interpreting 'Not Rejecting' the Null Hypothesis

Not rejecting the null hypothesis does not necessarily mean that we have proven the null hypothesis to be true. It simply suggests that, given the data and the chosen significance level, there is no strong evidence against the null hypothesis. The mean could still be exactly 100 nm, or it might differ but the difference is not statistically significant based on the collected data.
04

Conclusion on Evidence

Consequently, not rejecting the null hypothesis provides less evidence that the alternative hypothesis is true. However, it does not provide strong evidence in favor of the null hypothesis. It indicates that the critical mean width of 100 nm remains consistent with data but is not confirmed.

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Key Concepts

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

Understanding the Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a fundamental element of hypothesis testing. It is the initial assumption or claim that there is no effect or difference from a standard or baseline condition. In the context of our semiconductor example, the null hypothesis states that the critical dimension mean width of the component is exactly 100 nm.

Scientists and researchers set up the null hypothesis as a statement to be tested, not necessarily because they believe it to be true, but to serve as a basis for assessing evidence against it. If enough statistical evidence is found against the null hypothesis, it can be rejected in favor of an alternative hypothesis.

This is critical, as rejecting the null hypothesis can lead to new insights or changes in manufacturing processes, ensuring the results accurately reflect the tested conditions.
Exploring the Alternative Hypothesis
The alternative hypothesis, denoted \( H_1 \) or \( H_a \), is what you might conclude if the null hypothesis is found to be false. It is the hypothesis that indicates the presence of an effect or difference. For instance, in our semiconductor test, the alternative hypothesis could be that the mean width of the critical dimension is not equal to 100 nm. This would be expressed as \( H_1: \mu eq 100 \, \text{nm} \).

The alternative hypothesis is what researchers aim to support. It provides a new direction based on the data showing that something other than the initial assumption (the null hypothesis) is true.
  • It's vital to frame both the null and alternative hypotheses accurately,
  • as their clarity influences the interpretation of the test outcomes.
In hypothesis testing, failing to reject the null does not automatically support the alternative hypothesis; it simply means there is insufficient evidence to favor one over the other.
The Role of Statistical Evidence
Statistical evidence is the data and statistical analysis outcomes that help support or refute a hypothesis. In hypothesis testing, the strength of statistical evidence determines whether the null hypothesis can be rejected or not.

In our semiconductor example, the conclusion of not rejecting the null hypothesis is based on statistical evidence suggesting no significant deviation from a mean of 100 nm. However, it is important to understand the nuances: the absence of evidence is not evidence of absence. Here are some key points about statistical evidence in this context:
  • Not rejecting the null hypothesis is not equivalent to proving it.
  • Statistical evidence may be inconclusive, needing further testing or more data.
  • Confidence intervals provide a range where the true mean is likely found, giving context beyond the binary decision of hypothesis rejection or acceptance.
The takeaway is that statistical evidence fans the flames of inquiry and ensures decisions are backed by quantifiable data.
Understanding Significance Level
The significance level, typically denoted by \( \alpha \), is a threshold set by researchers which dictates the probability of rejecting a true null hypothesis. It reflects the level of risk they are willing to accept for a Type I error, which is the incorrect rejection of the null hypothesis.

Common significance levels are 0.05, 0.01, or 0.10, corresponding to 5%, 1%, or 10% risks, respectively. In practice, the significance level impacts the conclusion of a hypothesis test. For the semiconductor exercise, the significance level chosen would affect the decision to reject or not reject the null hypothesis that the mean width is 100 nm. If the computed p-value from the statistical test is less than or equal to the significance level, the null hypothesis is rejected.
  • It is crucial that the significance level is set before data collection and analysis begins to avoid biased conclusions.
  • This preset threshold helps quantify the certainty of the decision-making process in statistical analysis.
  • Ultimately, the significance level is a balancing act: minimizing both Type I and Type II errors while ensuring sufficiently strong evidence is present before rejecting a null hypothesis.
Understanding these nuances allows for a fair assessment of the hypothesis without overestimating the certainty of the results.

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Most popular questions from this chapter

Output from a software package is given below: One-Sample Z: Test of \(m u=20\) vs \(>20\) The assumed standard deviation \(=0.75\) $$ \begin{array}{lrrcccc} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\ \mathrm{x} & 10 & 19.889 & ? & 0.237 & ? & ? \end{array} $$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the above data to construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean. (d) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 20 ?\)

The proportion of residents in Phoenix favoring the building of toll roads to complete the freeway system is believed to be \(p=0.3\). If a random sample of 10 residents shows that 1 or fewer favor this proposal, we will conclude that \(p<0.3\). (a) Find the probability of type I error if the true proportion is \(p=0.3\) (b) Find the probability of committing a type II error with this procedure if \(p=0.2\) (c) What is the power of this procedure if the true proportion is \(p=0.2 ?\)

An article in Food Chemistry ["A Study of Factors Affecting Extraction of Peanut (Arachis Hypgaea L.) Solids with Water" (1991, Vol. 42, No. 2, pp. \(153-165\) ) ] found the percent protein extracted from peanut milk as follows: 78.3 \(\begin{array}{llllll}71.3 & 84.5 & 87.8 & 75.7 & 64.8 & 72.5\end{array}\) $$ \begin{array}{llllllll} 78.2 & 91.2 & 86.2 & 80.9 & 82.1 & 89.3 & 89.4 & 81.6 \end{array} $$ (a) Can you support a claim that mean percent protein extracted exceeds 80 percent? Use \(\alpha=0.05 .\) (b) Is there evidence that percent protein extracted is normally distributed? (c) What is the \(P\) -value of the test statistic computed in part (a)?

A random sample of 300 circuits generated 13 defectives. (a) Use the data to test \(H_{0}: p=0.05\) versus \(H_{1}: p \neq 0.05 .\) Use \(\alpha=0.05 .\) Find the \(P\) -value for the test. (b) Explain how the question in part (a) could be answered with a confidence interval.

The mean water temperature downstream from a power plant cooling tower discharge pipe should be no more than \(100^{\circ} \mathrm{F}\). Past experience has indicated that the standard deviation of temperature is \(2^{\circ} \mathrm{F}\). The water temperature is measured on nine randomly chosen days, and the average temperature is found to be \(98^{\circ} \mathrm{F}\). (a) Is there evidence that the water temperature is acceptable at \(\alpha=0.05 ?\) (b) What is the \(P\) -value for this test? (c) What is the probability of accepting the null hypothesis at \(\alpha=0.05\) if the water has a true mean temperature of \(104^{\circ} \mathrm{F} ?\)

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