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Chapter 9: Problem 123

The titanium content in an aircraft-grade alloy is an important determinant of strength. A sample of 20 test coupons reveals the following titanium content (in percent): \(8.32,8.05,8.93,8.65,8.25,8.46,8.52,8.35,8.36,8.41,8.42,\) \(8.30,8.71,8.75,8.60,8.83,8.50,8.38,8.29,8.46\) The median titanium content should be \(8.5 \%\). (a) 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 \(H_{0}: \tilde{\mu}=8.5\) versus \(H_{1}: \tilde{\mu} \neq 8.5\) with \(\alpha=0.05 .\) What is the \(P\) -value for this test?

Short Answer

Expert verified
For both tests, fail to reject the null hypothesis; median\( \tilde{\mu} \) likely 8.5.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis, \( H_0 \), is that the median titanium content is \( 8.5\% \). The alternative hypothesis, \( H_1 \), is that the median titanium content is not \( 8.5\% \):\[ H_0 : \tilde{\mu} = 8.5 \]\[ H_1 : \tilde{\mu} eq 8.5 \]
02

Perform the Sign Test

Count how many of the data points are above and below 8.5. Out of 20 test coupons, 13 have less than 8.5 and 7 have more than 8.5. Use \( B \), the count of values below the hypothesized median, in the sign test. The sign test is a non-parametric test which uses the binomial distribution.
03

Calculate the P-value for the Sign Test

Use the binomial distribution with \( n = 20 \) and \( p = 0.5 \) (since under the null hypothesis, each side of the median is equally likely).Calculate the P-value for \( B = 13 \), which counts the number of signs less than 8.5:\[ \text{Two-tailed P-value} = 2 \times P(X \geq 13) \text{ under } X \sim \text{Binomial}(n=20, p=0.5) \]Using a binomial table or calculator, the P-value is found to be approximately 0.115.
04

Decision Rule for the Sign Test

Since the P-value \( \approx 0.115 \) and \( \alpha = 0.05 \), we fail to reject the null hypothesis. There is not enough evidence to say that the median is different from 8.5 under this test.
05

Normal Approximation Setup

Using the normal approximation for a binomial distribution, first calculate the mean and standard deviation:\[ \mu = \frac{n}{2} = 10 \]\[ \sigma = \sqrt{\frac{n}{4}} = \sqrt{5} \approx 2.236 \]
06

Conduct the Normal Approximation Test

Convert the observed count \( B = 13 \) into a Z-score using the continuity correction:\[ Z = \frac{13 - 0.5 - 10}{\sqrt{5}} \approx \frac{2.5}{2.236} \approx 1.12 \]
07

Calculate P-value using Z-score

Determine the two-tailed P-value corresponding to \( Z = 1.12 \). The P-value is calculated as \( P(|Z| > 1.12) \). Using a Z-table or calculator, this gives: \[ P \approx 2(0.131) = 0.262 \]
08

Decision Rule for Normal Approximation

As the P-value is \( 0.262 \), which is greater than \( \alpha = 0.05 \), we again fail to reject the null hypothesis. We do not have sufficient evidence to conclude the median is different from 8.5.

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Key Concepts

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

Sign Test
The sign test is a simple non-parametric method for testing whether a sample median matches a hypothesized value. This test is useful when data doesn't meet the assumptions required for a t-test, such as normal distribution.
When applying the sign test, each observed value is compared to the hypothesized median. We note whether it is above or below it. If the hypothesized median is correct, the number of values above and below should be about equal if the sample size is sufficiently large.
For a two-tailed test like the one in this example, we test:
  • Null Hypothesis, \(H_0:\) The median is \(8.5\%\).
  • Alternative Hypothesis, \(H_1:\) The median is not \(8.5\%\).
By tallying how many numbers exceed or fall short of the median, we compute a statistic based on the binomial distribution.
Normal Approximation
Normal approximation provides a way to simplify hypothesis testing for a binomial distribution, especially if calculating directly from the binomial is complex or time-consuming.
When the sample size is sufficiently large, the binomial distribution can be approximated using a normal distribution. This is thanks to the Central Limit Theorem. The transition from binomial to normal involves calculating a Z-score.
To apply normal approximation to a sign test:
  • Calculate the mean \(\mu = \frac{n}{2}\).
  • Find the standard deviation \(\sigma = \sqrt{\frac{n}{4}}\).
For continuity correction, adjust the calculated statistic before converting it to a Z-score. This step is important when using a continuous normal distribution to approximate a discrete binomial distribution.
Binomial Distribution
The binomial distribution is fundamental in statistics for modeling binary outcomes, such as success or failure, in experiments repeated several times under the same conditions.
In the context of hypothesis testing, it helps in determining the probability of obtaining a certain number of "successes" in a sample of fixed size \(n\), with a specified probability \(p\) of success in a single trial.
For the sign test:
  • Each data point's placement above or below the hypothesized median is considered a success or failure.
  • The number of successes below or above this median is counted.
  • This forms the basis of determining how likely this outcome is, assuming the null hypothesis is true.
The binomial distribution facilitates the computation of the probability, aiding in the decision-making process during hypothesis testing.
P-value Calculation
The P-value quantifies statistical significance by offering the probability that the observed results (or more extreme results) occur by random chance, assuming the null hypothesis is true.
P-value calculation involves
  • Determining the tail-area probability.
  • In a sign test, this involves using a binomial distribution to find the probability of results as extreme as those observed.
  • Via normal approximation, using the Z-score to find this probability from the standard normal distribution.
Once calculated, the P-value is compared against a chosen significance level, \(\alpha\), often set at \(0.05\). If the P-value is less than \(\alpha\), the null hypothesis is usually rejected, indicating that the observed effect is statistically significant.

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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 \(0.9 ?\)

Patients in a hospital are classified as surgical or medical. A record is kept of the number of times patients require nursing service during the night and whether or not these patients are on Medicare. The data are presented here: $$\begin{array}{ccc} &{\text { Patient Category }} \\\\\hline \text { Medicare } & \text { Surgical } & \text { Medical } \\\\\text { Yes } & 46 & 52 \\\\\text { No } & 36 & 43\end{array}$$ Test the hypothesis (using \(\alpha=0.01\) ) that calls by surgicalmedical patients are independent of whether the patients are receiving Medicare. Find the \(P\) -value for this test.

An article in the ASCE Journal of Energy Engineering (1999, Vol. \(125,\) pp. \(59-75\) ) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures \(\left({ }^{\circ} \mathrm{C}\right)\) reported were as follows: \(23.01,22.22,22.04,22.62,\) and \(22.59 .\) (a) Test the hypotheses \(H_{0}: \mu=22.5\) versus \(H_{1}: \mu \neq 22.5,\) using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that interior temperature is normally distributed. (c) Compute the power of the test if the true mean interior temperature is as high as 22.75 . (d) What sample size would be required to detect a true mean interior temperature as high as 22.75 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 two-sided confidence interval on the mean interior temperature.

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 \(H_{1}: \mu \neq 99 ?\)

An article in Biological Trace Element Research \(\left[{ }^{\circ \cdot}\right.\) Interaction of Dietary Calcium, Manganese, and Manganese Source (Mn Oxide or Mn Methionine Complex) or Chick Performance and Manganese Utilization" (1991, Vol. 29(3), pp. 217-228)] showed the following results of tissue assay for liver manganese (ppm) in chicks fed high Ca diets. $$\begin{array}{llllll}6.02 & 6.08 & 7.11 & 5.73 & 5.32 & 7.10 \\\5.29 & 5.84 & 6.03 & 5.99 & 4.53 & 6.81\end{array}$$ (a) Test the hypothesis \(H_{0}: \sigma^{2}=0.6\) versus \(H_{1}: \sigma^{2} \neq 0.6\) using\(\alpha=0.01\) (b) What is the \(P\) -value for this test? (c) Discuss how part (a) could be answered by constructing a \(99 \%\) two-sided confidence interval for \(\sigma .\)
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