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Let the test statistic \(T\) have a \(t\) distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), df \(=15\), rejection region \(t \geq 3.733\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}, n=24\), rejection region \(t \leq-2.500\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}, n=31\), rejection region \(t \geq 1.697\) or \(t \leq\) \(-1.697\)

Short Answer

Expert verified
a) \( \alpha = 0.001 \); b) \( \alpha = 0.01 \); c) \( \alpha = 0.10 \).

Step by step solution

01

Understanding the significance level

The significance level \( \alpha \) represents the probability of rejecting the null hypothesis \( H_0 \) when it is actually true. In the context of a \( t \)-distribution, it corresponds to the probability in the tails of the distribution defined by the rejection region.
02

Find the significance level for case (a)

For \( H_a: \mu > \mu_0 \) with df = 15, and rejection region \( t \geq 3.733 \), we need to find \( P(T \geq 3.733) \) from the \( t \)-distribution table with 15 degrees of freedom. This probability is the significance level \( \alpha \). Lookup the table or use a statistical calculator to find this probability, which is \( \alpha = 0.001 \).
03

Find the significance level for case (b)

For \( H_a: \mu < \mu_0 \) with \( n = 24 \) (implying df = 23), the rejection region is \( t \leq -2.500 \). Calculate \( P(T \leq -2.500) \) with 23 degrees of freedom. Use a \( t \)-distribution table or calculator to find this probability, resulting in \( \alpha = 0.01 \).
04

Find the significance level for case (c)

For \( H_a: \mu eq \mu_0 \) with \( n = 31 \) (implying df = 30), the rejection regions are \( t \geq 1.697 \) or \( t \leq -1.697 \). Calculate \( P(T \geq 1.697) + P(T \leq -1.697) \) for 30 degrees of freedom. Each tail probability is \( 0.05/2 \), so the total is \( \alpha = 0.10 \).
05

Conclusion

Having calculated the different \( \alpha \) levels for each scenario using \( t \)-distribution tables or calculators, you can summarize them as the significance levels for the different alternate hypotheses.

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Key Concepts

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

Understanding Significance Level
The significance level, often represented as \( \alpha \), is a key concept in hypothesis testing. It helps us determine the probability of incorrectly rejecting the null hypothesis \( H_0 \) when it is actually true. This is also known as making a Type I error. The significance level sets a threshold for judging whether the results of an experiment are statistically significant or not. In the context of a \( t \)-distribution, the significance level corresponds to the probability encompassed by the rejection region in the tails of the distribution. For example, if \( \alpha = 0.05 \), this means there is a 5% probability of rejecting \( H_0 \) incorrectly.
  • Lower significance levels imply stricter criteria for rejecting \( H_0 \), reducing the risk of Type I errors.
  • Common \( \alpha \) values are 0.05, 0.01, and 0.10.
It is important for researchers to choose an appropriate significance level based on their specific scientific or practical context before conducting the test.
Decoding Degrees of Freedom
Degrees of freedom, abbreviated as df, are crucial for understanding statistical tests like the \( t \)-distribution. They refer to the number of independent values or quantities that can vary in a statistical calculation. The concept applies to numerous statistical tests, where it helps determine the specific form and critical values from the distribution being utilized.
In the context of the \( t \)-distribution:
  • The degrees of freedom can be calculated as \( n-1 \) for a single sample, where \( n \) is the sample size.
  • More degrees of freedom typically mean the \( t \)-distribution resembles the standard normal distribution closer.
Accurate calculation of the degrees of freedom is essential because it affects the shape of the \( t \)-distribution and thus the critical values used in determining significance levels. Always pay close attention to determining the correct degrees of freedom for accurate test results.
Exploring the Rejection Region
The rejection region is a fundamental concept in hypothesis testing that directly connects with the significance level and the decision to reject the null hypothesis \( H_0 \). It represents the range of values that leads us to reject \( H_0 \) in favor of the alternative hypothesis \( H_a \). The rejection region lies in the tails of the probability distribution of the test statistic. Calculating the rejection region involves using the critical values obtained from the distribution, which in this context is a \( t \)-distribution.
Key points regarding the rejection region:
  • A one-tailed test, like \( H_a: \mu > \mu_0 \), will have its rejection region on one side of the distribution.
  • A two-tailed test, such as \( H_a: \mu eq \mu_0 \), will have rejection regions at both extremes of the distribution.
  • Width and location of the rejection region depend on the chosen significance level.
By precisely identifying this region, researchers can discern when their test results are significant, safely making conclusions about their hypotheses.
Demystifying Hypothesis Testing
Hypothesis testing is a cornerstone of statistical analysis, allowing researchers to draw inferences about population parameters based on sample data. This process begins with formulating two competing hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \).
The primary aim in hypothesis testing is to determine sufficient evidence in the sample to reject \( H_0 \). Here’s how the process unfolds:
  • Determine a significance level \( \alpha \), which defines the probability threshold for rejecting \( H_0 \).
  • Choose the appropriate statistical test, such as a \( t \)-test when dealing with small sample sizes and unknown population standard deviations.
  • Calculate the test statistic and compare it against the critical values from the distribution corresponding to \( H_0 \).
  • Based on this comparison, decide whether to reject or fail to reject \( H_0 \).
By following this structured framework, hypothesis testing helps researchers make data-driven decisions, providing insights into the reliability and implications of their data.

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Most popular questions from this chapter

A sample of 50 lenses used in eyeglasses yields a sample mean thickness of \(3.05 \mathrm{~mm}\) and a sample standard deviation of \(.34 \mathrm{~mm}\). The desired true average thickness of such lenses is \(3.20 \mathrm{~mm}\). Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using \(\alpha=.05\).

It is known that roughly \(2 / 3\) of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? The article "Human Behavior: Adult Persistence of Head-Turning Asymmetry" (Nature, 2003: 771) reported that in a random sample of 124 kissing couples, both people in 80 of the couples tended to lean more to the right than to the left. a. If \(2 / 3\) of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 124 who do so differs from the expected value by at least as much as what was actually observed? b. Does the result of the experiment suggest that the \(2 / 3\) figure is implausible for kissing behavior? State and test the appropriate hypotheses.

The sample average unrestrained compressive strength for 45 specimens of a particular type of brick was computed to be \(3107 \mathrm{psi}\), and the sample standard deviation was 188 . The distribution of unrestrained compressive strength may be somewhat skewed. Does the data strongly indicate that the true average unrestrained compressive strength is less than the design value of 3200 ? Test using \(\alpha=.001\).

The calibration of a scale is to be checked by weighing a 10 -kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either \(\bar{x} \geq\) \(10.1032\) or \(\bar{x} \leq 9.8968\). What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judged unnecessary when in fact \(\mu=10.1\) ? When \(\mu=9.8\) ? d. Let \(z=(\bar{x}-10) /(\sigma / \sqrt{n})\). For what value \(c\) is the rejection region of part (b) equivalent to the "two-tailed" region either \(z \geq c\) or \(z \leq-c\) ? e. If the sample size were only 10 rather than 25 , how should the procedure of part (d) be altered so that \(\alpha=.05\) ? f. Using the test of part (e), what would you conclude from the following sample data: $$ \begin{array}{rrrrr} 9.981 & 10.006 & 9.857 & 10.107 & 9.888 \\ 9.728 & 10.439 & 10.214 & 10.190 & 9.793 \end{array} $$ g. Reexpress the test procedure of part (b) in terms of the standardized test statistic \(Z=(\bar{X}-10) /(\sigma / \sqrt{n})\).

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Patterns of Older Americans: A National Study" (J. Gerontology, 1992: M145-150) reports the following summary data on intake for a sample of males age \(65-74\) years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance?

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