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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
No, not rejecting the null hypothesis isn't strong evidence the mean equals 100 nm.

Step by step solution

01

Understanding Hypothesis Testing

In hypothesis testing, we start with a null hypothesis ( H_0 ) and an alternative hypothesis ( H_1 ). The null hypothesis in this case is that the mean width of the critical dimension is 100 ext{ nm} .
02

Interpreting 'Not Rejecting the Null Hypothesis'

Not rejecting H_0 means there wasn't enough statistical evidence to support the alternative hypothesis, which would be that the mean is different from 100 ext{ nm} . However, this does not confirm that 100 ext{ nm} is the actual mean.
03

Explaining Lack of Strong Evidence

The decision to not reject H_0 doesn't prove H_0 is true. It only shows that there isn't sufficient evidence against H_0 . Limitations such as sample size and variability could affect this test result, meaning that the true mean could be different from 100 ext{ nm} but isn't detected due to these factors.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often symbolized as \(H_0\), is a foundational concept in statistical hypothesis testing. It presents a statement we seek to test and potentially reject. In the context of the semiconductor manufacturer's data, the null hypothesis was that the mean width of a critical dimension is exactly \(100\) nm. The idea is to assume this statement is true until evidence indicates otherwise.

When you perform hypothesis testing, you develop two competing hypotheses: the null hypothesis (\(H_0\)), which posits no change or effect, and the alternative hypothesis (\(H_1\)), which suggests a deviation from \(H_0\). In hypothesis testing, the aim is to determine if the sample data provides enough evidence to reject \(H_0\) in favor of \(H_1\).

An important point to remember is that failing to reject \(H_0\) doesn't mean we accept it as true. It simply means that there wasn't enough statistical evidence in the sample to conclude that \(H_1\) is more plausible than \(H_0\).
Statistical Evidence
Statistical evidence is the information collected from data, which is used to assess the validity of the null hypothesis. When data doesn't provide strong evidence against the null hypothesis, we do not reject it. This is what happened in the semiconductor manufacturer's case. Simply put, the test's results were not convincing enough to argue that the mean width is anything other than \(100\) nm.

It's crucial to understand that lack of evidence against \(H_0\) doesn't prove \(H_0\) true. It only shows that the data didn't refute it. Statistical tests often employ a significance level, typically \(\alpha = 0.05\), which helps determine if the results are statistically significant. A p-value is calculated and then compared against \(\alpha\) to decide if \(H_0\) should be rejected.
  • If the p-value is less than \(\alpha\), \(H_0\) is rejected.
  • If the p-value is greater than \(\alpha\), we do not reject \(H_0\).
Therefore, the conclusion from the test to "not reject \(H_0\)" suggests a lack of strong statistical evidence, but not proof of the null hypothesis.
Sample Size Influence
Sample size plays a critical role in hypothesis testing and the strength of the evidence gathered. It's a pivotal factor that affects the power of a statistical test. The power of a test is the probability of correctly rejecting a false \(H_0\).

In our scenario, a small sample size might not adequately represent the population, potentially leading to incorrect conclusions about \(H_0\). A larger sample size could provide more reliable evidence, increasing the likelihood of detecting a true effect if one exists.

Here are some reasons why sample size matters:
  • **Greater Precision:** A larger sample size reduces the margin of error, providing more accurate estimates of the population parameter.
  • **Increased Test Power:** More data enhances the ability to detect deviances from \(H_0\), reducing the risk of Type II error (failing to reject a false \(H_0\)).
  • **Stability in Results:** Larger samples tend to yield consistent and stable results.
Thus, if the sample size was inadequate in our test, we might have insufficient evidence to make a definitive conclusion about the critical dimension mean.

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Most popular questions from this chapter

Suppose that 10 sets of hypotheses of the form $$H_{0}: \mu=\mu_{0} \quad H_{1}: \mu \neq \mu_{0}$$ have been tested and that the \(P\) -values for these tests are 0.12 , \(0.08 .0 .93,0.02,0.01,0.05,0.88,0.15,0.13,\) and \(0.06 .\) Use Fisher's procedure to combine all of these \(P\) -values. What conclusions can you draw about these hypotheses?

Suppose that we wish to test the hypothesis \(H_{0}: \mu=85\) versus the alternative \(H_{1}: \mu>85\) where \(\sigma=16\). Suppose that the true mean is \(\mu=86\) and that in the practical context of the problem, this is not a departure from \(\mu_{0}=85\) that has practical significance. (a) For a test with \(\alpha=0.01,\) compute \(\beta\) for the sample sizes \(n=\) \(25,100,400,\) and 2500 assuming that \(\mu=86 .\) (b) Suppose that the sample average is \(\bar{x}=86 .\) Find the \(P\) -value for the test statistic for the different sample sizes specified in part (a). Would the data be statistically significant at \(\alpha=0.01 ?\) (c) Comment on the use of a large sample size in this problem.

Consider the computer output below Test of \(p=0.25\) vs \(p<0.25\) $$\begin{array}{cccccc}\mathrm{X} & \mathrm{N} & \text { Sample } p & \text { Bound } & Z \text { -Value } & P \text { -Value } \\\53 & 225 & 0.235556 & 0.282088 & ? & ?\end{array}$$ Using the normal approximation: (a) Fill in the missing information. (b) What are your conclusions if \(\alpha=0.05 ?\) (c) The normal approximation to the binomial was used here. Was that appropriate? (d) Find a \(95 \%\) upper-confidence bound on the true proportion. (e) What are the \(P\) -value and your conclusions if the alternative hypothesis is \(H_{1}: p \neq 0.25 ?\)

Output from a software package follows: Test of \(m u=99\) vs \(>99\) The assumed standard deviation \(=2.5\) $$\begin{array}{ccccccc}\text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\\x & 12 & 100.039 & 2.365 & ? & 1.44 & 0.075 \\\\\hline\end{array}$$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) If the hypothesis had been \(H_{0}: \mu=98\) versus \(H_{0}: \mu>98\), would you reject the null hypothesis at the 0.05 level of significance? Can you answer this without referring to the normal table? (d) Use the normal table and the preceding data to construct a \(95 \%\) lower bound on the mean. (e) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 99 ?\)

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels yearround to reduce air pollution. If more than 315 voters respond positively, we will conclude that at least \(60 \%\) of the voters favor the use of these fuels. (a) Find the probability of type I error if exactly \(60 \%\) of the voters favor the use of these fuels. (b) What is the type II error probability \(\beta\) if \(75 \%\) of the voters favor this action?

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