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The true average diameter of ball bearings of a certain type is supposed to be \(0.5\) in. What conclusion is appropriate when testing \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5 \mathrm{in}\) each of the following situations: a. \(n=13, t=1.6, \alpha=.05\) b. \(n=13, t=-1.6, \alpha=.05\) c. \(n=25, t=-2.6, \alpha=.01\) d. \(n=25, t=-3.6\)

Short Answer

Expert verified
In case a and b we fail to reject the null hypothesis while in case c we reject the null hypothesis. In case d, there is no definitive answer without a significance level but the t-value indicates it might reject the null hypothesis.

Step by step solution

01

Determine the critical t-value

For situation a. Using a t-table or t-distribution calculator, with \(n = 13\) thus degrees of freedom equals \(n-1 = 12\) and \(\alpha = .05\), we get a critical t-value of approximately ±2.18; For situation b. Like in case a, we get the same critical t-value which is approximately ±2.18; For situation c. In this case, with \(n=25\), thus degrees of freedom equals \(n-1 = 24\) and \(\alpha = .01\), we get a critical t-value of approximately ±2.80; For situation d. In absence of a significance level, we can only infer a conclusion based on the t-value alone.
02

Compare the t-values

For situation a. Given t-value is \(1.6\), which is less than the critical t-value \(|2.18|\). Therefore, we fail to reject the null hypothesis; For situation b. Given t-value is \(-1.6\), which again is within the range of \(-2.18\) and \(2.18\). Therefore, we fail to reject the null hypothesis; For situation c. Given t-value is \(-2.6\), which is less than the lower limit of the critical range \(-2.8\). Therefore, we reject the null hypothesis; For situation d. We cannot make definitive conclusion but the provided value \(-3.6\) suggests that it might reject the null hypothesis if we had a significance level.

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Key Concepts

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

Understanding t-distribution
The t-distribution is a critical component in hypothesis testing, especially useful when dealing with smaller sample sizes. It represents the distribution of sample means drawn from a normally distributed population when the population standard deviation is unknown. This distribution is symmetrical and more spread out compared to the standard normal distribution, which makes it better suited for small datasets.

A key aspect of the t-distribution involves "degrees of freedom," calculated as the sample size minus one. This number helps determine the shape of the t-distribution. The fewer the degrees of freedom, the wider and more variegated the distribution becomes, leading to larger critical values. As the sample size increases, the t-distribution approaches a normal distribution. Understanding this behavior is crucial for analyzing small sample data effectively.
The role of the critical t-value
The critical t-value is a threshold that helps determine whether an observed t-statistic suggests significant evidence against the null hypothesis. This value is derived from the t-distribution chart or calculator, based on the degrees of freedom and the desired significance level.

For example, in a hypothesis test where the null hypothesis states the population mean is 0.5 inches, the critical t-value will dictate the range within which we would fail to reject the null hypothesis. If your calculated t-value from the sample data falls outside of this range, it signifies that the sample provides enough evidence to reject the null hypothesis. It acts as a cutoff point that marks the significance boundary for making decisions in hypothesis testing.
Understanding the null hypothesis
The null hypothesis (\(H_{0}\)) is a fundamental part of hypothesis testing. It is a statement that asserts no effect or no difference exists. In our example, the null hypothesis states that the average diameter of ball bearings is exactly 0.5 inches.

When conducting a hypothesis test, the goal is to gather evidence to decide whether to reject this null hypothesis in favor of an alternative hypothesis (\(H_{a}\)), which in this case is that the average diameter is not equal to 0.5 inches.
  • The process involves computing a sample statistic (such as a t-value)
  • Comparing it to a critical value derived from a relevant distribution
Failing to reject the null hypothesis doesn't prove it true; rather, it suggests that the sample data isn’t sufficient to demonstrate a significant difference.
The significance level in hypothesis testing
The significance level, often denoted as \(\alpha\), is a threshold set by the researcher to determine how extreme the sample data must be in order to reject the null hypothesis. Common significance levels are 0.05, 0.01, or 0.10. These numbers represent the risk of committing a Type I error, which is rejecting a true null hypothesis.

Choosing a significance level is essential in hypothesis testing as it balances the risk of false positives. For instance, an \(\alpha\) of 0.05 implies a 5% risk that the null hypothesis will be incorrectly rejected. It provides a boundary for how unlikely your results must be (assuming the null hypothesis is true) to consider them statistically significant.

So, in a hypothesis test examination such as the ball bearing example, the significance level helps quantify the uncertainty in drawing conclusions about the population.

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Most popular questions from this chapter

A county commissioner must vote on a resolution that would commit substantial resources to the construction of a sewer in an outlying residential area. Her fiscal decisions have been criticized in the past, so she decides to take a survey of constituents to find out whether they favor spending money for a sewer system. She will vote to appropriate funds only if she can be fairly certain that a majority of the people in her district favor the measure. What hypotheses should she test?

What motivates companies to offer stock ownership plans to their employees? In a random sample of 87 companies having such plans, 54 said that the primary rationale was tax related ("The Advantages and Disadvantages of ESOPs: A Long- Range Analysis," Journal of Small Business Management [1991]: \(15-21\) ). Does this information provide strong support for concluding that more than half of all such firms feel this way?

An automobile manufacturer who wishes to advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon) decides to carry out a fuel efficiency test. Six nonprofessional drivers are selected, and each one drives a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{llllll}27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6\end{array}\) Assuming that fuel efficiency is normally distributed under these circumstances, do the data contradict the claim that true average fuel efficiency is (at least) \(30 \mathrm{mpg}\) ?

The amount of shaft wear after a fixed mileage was determined for each of 7 randomly selected internal combustion engines, resulting in a mean of \(0.0372\) in. and a standard deviation of \(0.0125 \mathrm{in}\). a. Assuming that the distribution of shaft wear is normal, test at level \(.05\) the hypotheses \(H_{0}: \mu=.035\) versus \(H_{a}:\) \(\mu>.035\) b. Using \(\sigma=0.0125, \alpha=.05\), and Appendix Table 5 . what is the approximate value of \(\beta\), the probability of a Type II error, when \(\mu=.04\) ? c. What is the approximate power of the test when \(\mu=\) \(.04\) and \(\alpha=.05\) ?

Let \(\mu\) denote the true average diameter for bearings of a certain type. A test of \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) will be based on a sample of \(n\) bearings. The diameter distribution is believed to be normal. Determine the value of \(\beta\) in each of the following cases: a. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.52\) b. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.48\) c. \(n=15, \alpha=.01, \sigma=0.02, \mu=0.52\) d. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.54\) e. \(n=15, \alpha=.05, \sigma=0.04, \mu=0.54\) f. \(n=20, \alpha=.05, \sigma=0.04, \mu=0.54\) \(\mathrm{g}\). Is the way in which \(\beta\) changes as \(n, \alpha, \sigma\), and \(\mu\) vary consistent with your intuition? Explain.

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