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A manufacturer produces crankshafts for an automobile engine. The wear of the crankshaft 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 \(0.9 ?\)

Short Answer

Expert verified
(a) Fail to reject \(H_0\); (b) Power is approximately 0.35; (c) Sample size required is about 45.

Step by step solution

01

Define the Null and Alternative Hypotheses

First, let's define the null hypothesis \(H_0:\ \mu = 3\) and the alternative hypothesis \(H_1:\ \mu eq 3\). This is a two-tailed test because the alternative hypothesis specifies not equal to \(3\).
02

Calculate the Test Statistic

To calculate the test statistic, use the formula for the Z-test, because the population standard deviation \(\sigma\) is known:\[Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Substitute the given values:\[\bar{x} = 2.78, \mu = 3, \sigma = 0.9, n = 15\]\[Z = \frac{2.78 - 3}{\frac{0.9}{\sqrt{15}}} = \frac{-0.22}{0.2324} \approx -0.947\]
03

Determine the Critical Z-values

For a significance level \(\alpha = 0.05\) in a two-tailed test, the critical Z-values are \(\pm 1.96\). If the calculated test statistic falls outside this range, we reject the null hypothesis.
04

Make a Decision about the Null Hypothesis

Since the calculated Z-statistic \(-0.947\) does not fall outside the critical values of \(-1.96\) and \(1.96\), we fail to reject the null hypothesis. Thus, there is not enough evidence to suggest that the mean wear is different from 3.
05

Calculate the Power of the Test for \(\mu = 3.25\)

First, calculate the effect size:\[effect\ size = \frac{3.25 - 3}{0.9} = \frac{0.25}{0.9} = 0.278\]Then, calculate the non-centrality parameter:\[non\_centrality\_parameter = \frac{n\times effect\ size}{\sqrt{n}} = \frac{15 \times 0.278}{\sqrt{15}} \approx 1.073\]The power is calculated using statistical software or Z-tables considering the non-central parameter, resulting in power \(\approx 0.35\).
06

Determine the Sample Size for Desired Power

To achieve a power of at least 0.9 to detect a mean of 3.75, we setup:\[effect\ size = \frac{3.75 - 3}{0.9} = \frac{0.75}{0.9} \approx 0.833\]Using statistical power tables or software for \(\alpha = 0.05\), one-tailed, the sample size required to achieve 0.9 power will be approximately 45 when using typical calculations.

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Key Concepts

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

Z-test
The Z-test is a statistical method used to determine whether there is a significant difference between the sample mean and the population mean. In the scenario where the population standard deviation \( \sigma \) is known and the sample size (\( n \)) is sufficiently large, the Z-test is widely applicable. The Z-test formula is:
  • \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
Here, \( \bar{x} \) represents the sample mean, \( \mu \) is the hypothesized population mean, and \( n \) is the sample size. The Z-test gives us the test statistic, which we compare to critical values from the Z-distribution. This comparison helps us decide whether to accept or reject the null hypothesis. The Z-test is particularly useful when the sample size is large and the data follows a normal distribution, making it a powerful tool in hypothesis testing.
Significance Level
In hypothesis testing, the significance level \( \alpha \) represents the probability of rejecting a true null hypothesis, also known as the Type I error rate. It is a critical threshold that determines the rejection region for the hypothesis test. Commonly used significance levels include 0.05, 0.01, and 0.1, with \( \alpha = 0.05 \) often being the standard choice in many fields. When conducting a two-tailed test, as in the example provided, the significance level is equally split between the two tails of the normal distribution. If the calculated test statistic falls beyond this threshold, we reject the null hypothesis in favor of the alternative hypothesis. Thus, the significance level is essential for determining the critical values and helps us control the rate of false positives.
Power of a Test
The power of a test is the probability that the test correctly rejects a false null hypothesis. It is a measure of the test’s ability to detect an effect when it exists. High power is desirable because it indicates a greater chance of detecting true differences. Power is affected by the significance level, sample size, and the true difference from the null hypothesis. In practice, to calculate the power of a test, we use the non-centrality parameter or specialized statistical software. With a pre-specified effect size, power tells us the likelihood of observing an effect, assuming the true mean differs from the hypothesized mean. Generally, a power of 0.8 or higher is considered adequate for most tests, indicating an 80% chance of correctly detecting an effect.
Sample Size Calculation
Calculating the appropriate sample size is vital to ensure a hypothesis test has enough power to detect a meaningful effect. The required sample size depends on several factors: the significance level, the desired power, and the expected effect size.The effect size can be estimated as:
  • \[ effect\_size = \frac{\text{difference in means}}{\sigma} \]
Where \( \sigma \) is the standard deviation of the population. Once the effect size is known, statistical power tables or software can determine the sample size needed to achieve the desired power. For example, to achieve a power of 0.9 in detecting a hypothetical mean of 3.75, the sample size calculated might be roughly 45 under conventional settings. Proper sample size calculation ensures the test is neither underpowered (too small) nor wasteful (too large).

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Most popular questions from this chapter

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson random variables, each with parameter \(\lambda,\) and \(n\) is large, the sample mean \(\bar{X}\) has an approximate normal distribution with mean \(\lambda\) and variance \(\lambda / n\). Therefore, $$ Z=\frac{\bar{X}-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. Thus we can test \(H_{0}: \lambda=\lambda_{0}\) by replacing \(\lambda\) in \(Z\) by \(\lambda_{0}\). When \(X_{i}\) are Poisson variables, this test is preferable to the largesample test of Section \(9-2.3,\) which would use \(S / \sqrt{n}\) in the denominator, because it is designed just for the Poisson distribution. Suppose that the number of open circuits on a semiconductor wafer has a Poisson distribution. Test data for 500 wafers indicate a total of 1038 opens. Using \(\alpha=0.05,\) does this suggest that the mean number of open circuits per wafer exceeds \(2.0 ?\)

For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

The sodium content of twenty 300 -gram boxes of organic cornflakes was determined. The data (in milligrams) are as follows: \(131.15,130.69,130.91,129.54,129.64,128.77,\) 130.72,128.33,128.24,129.65,130.14,129.29,128.71,129.00 \(129.39,130.42,129.53,130.12,129.78,130.92 .\) (a) Can you support a claim that mean sodium content of this brand of cornflakes differs from 130 milligrams? Use \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check that sodium content is normally distributed. (c) Compute the power of the test if the true mean sodium content is 130.5 milligrams. (d) What sample size would be required to detect a true mean sodium content of 130.1 milligrams if we wanted the power of the test to be at least \(0.75 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean sodium content.

A hypothesis will be used to test that a population mean equals 7 against the alternative that the population mean does not equal 7 with unknown variance \(\sigma\). What are the critical values for the test statistic \(T_{0}\) for the following significance levels and sample sizes? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

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): $$ \begin{array}{l} 8.32,8.05,8.93,8.65,8.25,8.46,8.52,8.35,8.36,8.41,8.42, \\ 8.30,8.71,8.75,8.60,8.83,8.50,8.38,8.29,8.46 \end{array} $$ 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 \(H_{0}: \widetilde{\mu}=8.5\) versus \(H_{1}: \tilde{\mu} \neq 8.5,\) with \(\alpha=0.05 .\) What is the \(P\) -value for this test?

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