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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?

Short Answer

Expert verified
Reject the null hypothesis; significant evidence suggests deviations from \( \mu = \mu_0 \).

Step by step solution

01

Overview of Fisher's Method

Fisher's method is used to combine multiple independent p-values from different tests of the same hypothesis into a single test statistic to evaluate the overall significance. It is especially useful when dealing with multiple testing scenarios, allowing for determining the collective significance of test results.
02

Calculation of Fisher's Test Statistic

The Fisher's test statistic, denoted as \( X^2 \), is calculated using the formula: \[ X^2 = -2 \sum \ln(p_i) \]where \( p_i \) is each individual p-value. For our data: - \( p_1 = 0.12 \), \( p_2 = 0.08 \), \( p_3 = 0.93 \), \( p_4 = 0.02 \), \( p_5 = 0.01 \)- \( p_6 = 0.05 \), \( p_7 = 0.88 \), \( p_8 = 0.15 \), \( p_9 = 0.13 \), \( p_{10} = 0.06 \).Plug these values into the formula to evaluate \( X^2 \).
03

Computing Logarithms

Calculate the natural logarithm of each p-value:- \( \ln(0.12) \approx -2.12 \) - \( \ln(0.08) \approx -2.53 \)- \( \ln(0.93) \approx -0.07 \)- \( \ln(0.02) \approx -3.91 \)- \( \ln(0.01) \approx -4.61 \) - \( \ln(0.05) \approx -3.00 \)- \( \ln(0.88) \approx -0.13 \)- \( \ln(0.15) \approx -1.90 \)- \( \ln(0.13) \approx -2.04 \)- \( \ln(0.06) \approx -2.81 \).Add these values.
04

Summation of Logarithms

Sum up the logarithmic values obtained:\[ -2.12 + (-2.53) + (-0.07) + (-3.91) + (-4.61) + (-3.00) + (-0.13) + (-1.90) + (-2.04) + (-2.81) = -23.12 \]
05

Calculate Fisher's Statistic

Using \( X^2 = -2 \sum \ln(p_i) \) and substituting \( \sum \ln(p_i) = -23.12 \), compute:\[ X^2 = -2 \times (-23.12) = 46.24 \].
06

Determine Critical Value

The combined test statistic \( X^2 \) follows a \( \chi^2 \) distribution with degrees of freedom equal to twice the number of tests (\( 2 \times 10 = 20 \)). Compare \( X^2 = 46.24 \) to the critical value from \( \chi^2_{0.05} \) with 20 degrees of freedom, which is approximately 31.41.
07

Conclusion

Since the calculated \( X^2 = 46.24 \) exceeds the critical value of 31.41, we reject the null hypothesis \( H_0 \). This suggests that at least some of these hypotheses indicate a significant departure from the null hypothesis across the tests.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing involves making assumptions and determining if the available data supports these assumptions. At its core, it helps researchers decide whether observed effects are due to chance or if they point to something significant. The process begins with formulating a null hypothesis Assuming there is no effect or difference. To challenge this, we propose an alternative hypothesis with a supposed effect or difference. During testing, data is analyzed to measure their consistency with the null hypothesis. Test statistics are employed, and if the computed value is compellingly against the null hypothesis, the alternative hypothesis gains credence. However, the conclusion is probabilistic, termed statistically significant, rather than certain. Key steps include:
  • Formulate null and alternative hypotheses.
  • Collect and analyze relevant data.
  • Calculate appropriate test statistics.
  • Decide on hypothesis validity based on pre-defined thresholds.
Understanding this foundation is crucial as it underpins the usage of p-values and statistical significance in statistical analyses.
The Role and Interpretation of P-values
The p-value plays a central role in hypothesis testing by quantifying the evidence against a null hypothesis. Essentially, it measures the probability of observing results as extreme as, or more extreme than, the actual observations, assuming that the null hypothesis is true. The smaller the p-value, the stronger the evidence against the null hypothesis, suggesting that the alternative hypothesis might be more credible. Scientists often use thresholds like 0.05 or 0.01 as benchmarks for significance. If a p-value falls below these thresholds, the results are typically deemed statistically significant, meaning it's unlikely the observed effect is due to random chance.
However, it’s essential to remember:
  • A small p-value does not prove the alternative hypothesis; it indicates data inconsistency with the null hypothesis.
  • P-values do not measure the magnitude or importance of an effect, only presence versus absence.
  • P-values are dependent on sample size; large samples may produce small p-values even with trivial differences.
Hence, while p-values are crucial for inference, they must be complemented by other information like effect size to provide substantive insights.
Exploring the Chi-Squared Distribution
When dealing with Fisher's Method or similar approaches, understanding the chi-squared distribution is vital. This distribution is used for hypothesis tests that consider variance, especially in Fisher's Method, which involves combining p-values. In essence, the chi-squared distribution is the sum of the squares of standard normal variables.
It arises in tests of independence and goodness-of-fit, among others. Key characteristics are:
  • It is a family of distributions, with each characterized by its degrees of freedom (df).
  • Its shape depends on the degrees of freedom: skewed for low df, approximating normality as df increases.
  • Fisher's method utilizes it to determine if combined test significance exists across independent tests.
In the context of Fisher’s procedure, degrees of freedom play a critical role in establishing significance cutoffs. Understanding this helps interpret the results from combining p-values accurately.
Meaning and Importance of Statistical Significance
Statistical significance is a key concept in determining whether results from data analyses are likely due to specific effects rather than by random chance. In practical terms, when a result is statistically significant, it suggests that the observed finding is unlikely to have arisen purely as a result of variability in the sample data. Interpreting statistical significance involves comparing the p-value against a predetermined significance level (commonly 0.05 ). If the p-value is less than this threshold, the result is statistically significant, implying that the null hypothesis can be rejected. However, true understanding and utility stem from remembering that:
  • Significance does not imply importance. A statistically significant outcome could be due to a very small effect size.
  • Significance levels are somewhat arbitrary; they reflect conventional threshold choices rather than ironclad rules.
  • It is one piece of the puzzle and should be considered alongside confidence intervals and domain-specific context.
These realms of knowledge allow analysts and researchers to properly discern and interpret the meaning behind their statistical findings.

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Most popular questions from this chapter

A manufacturer produces crankshafts for an automobile engine. The crankshafts wear after 100,000 miles \((0.0001\) inch \()\) is of interest because it is likely to have an impact on warranty claims. A random sample of \(n=15\) shafts is tested and \(\bar{x}=2.78\). It is known that \(\sigma=0.9\) and that wear is normally distributed. a. Test \(H_{0}: \mu=3\) versus \(H_{1}: \mu \neq 3\) using \(\alpha=0.05\). b. What is the power of this test if \(\mu=3.25 ?\) c. What sample size would be required to detect a true mean of 3.75 if we wanted the power to be at least \(0.9 ?\)

Cloud seeding has been studied for many decades as a weather modification procedure (for an interesting study of this subject, see the article in Technometrics, "A Bayesian Analysis of a Multiplicative Treatment Effect in Weather Modification," \(1975,\) Vol. \(17,\) pp. \(161-166) .\) The rainfall in acre-feet from 20 clouds that were selected at random and seeded with silver nitrate follows: \(18.0,30.7,19.8,27.1,22.3,18.8,31.8,23.4,\) 21.2,27.9,31.9,27.1,25.0,24.7,26.9,21.8,29.2,34.8,26.7 and 31.6 a. Can you support a claim that mean rainfall from seeded clouds exceeds 25 acre-feet? Use \(\alpha=0.01 .\) Find the \(P\) -value. b. Check that rainfall is normally distributed. c. Compute the power of the test if the true mean rainfall is 27 acre-feet. d. What sample size would be required to detect a true mean rainfall of 27.5 acre-feet if you wanted the power of the test to be at least \(0.9 ?\) e. Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean diameter.

Ten samples were taken from a plating bath used in an electronics manufacturing process, and the \(\mathrm{pH}\) of the bath was determined. The sample \(\mathrm{pH}\) values are 7.91,7.85,6.82,8.01 \(7.46,6.95,7.05,7.35,7.25,\) and 7.42. Manufacturing engineering believes that \(\mathrm{pH}\) has a median value of \(7.0 .\) a. Do the sample data indicate that this statement is correct? Use the sign test with \(\alpha=0.05\) to investigate this hypothesis. Find the \(P\) -value for this test. b. Use the normal approximation for the sign test to test \(H_{0}: \tilde{\mu}=7.0\) versus \(H_{0}: \tilde{\mu} \neq 7.0 .\) What is the \(P\) -value for this test?

An article in the British Medical Journal ["Comparison of Treatment of Renal Calculi by Operative Surgery, Percutaneous Nephrolithotomy, and Extracorporeal Shock Wave Lithotripsy" (1986, Vol. 292, pp. \(879-882\) ) ] reported that percutaneous nephrolithotomy (PN) had a success rate in removing kidney stones of 289 out of 350 (83\%) patients. However, when the stone diameter was considered, the results looked different. For stones of \(<2 \mathrm{~cm}, 87 \%(234 / 270)\) of cases were successful. For stones of \(\geq 2 \mathrm{~cm},\) a success rate of \(69 \%(55 / 80)\) was observed for PN. a. Are the successes and size of stones independent? Use \(\alpha=0.05\) b. Find the \(P\) -value for this test.

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