/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 In a random sample of 85 automob... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish roughness that exceeds the specifications. Do these data present strong evidence that the proportion of crankshaft bearings exhibiting excess surface roughness exceeds \(0.10 ?\) (a) State and test the appropriate hypotheses using \(\alpha=0.05\). (b) If it is really the situation that \(p=0.15,\) how likely is it that the test procedure in part (a) will not reject the null hypothesis? (c) If \(p=0.15,\) how large would the sample size have to be for us to have a probability of correctly rejecting the null hypothesis of \(0.9 ?\)

Short Answer

Expert verified
(a) Fail to reject the null hypothesis; (b) \( \beta \approx 0.242 \); (c) Required sample size is 238.

Step by step solution

01

Define the Hypotheses

We want to determine if the proportion of crankshaft bearings exhibiting excess surface roughness exceeds 0.10. Therefore, the null and alternative hypotheses are:\[ H_0: p = 0.10 \]\[ H_a: p > 0.10 \]
02

Check Normality Condition and Calculate Sample Proportion

First, verify the normal approximation condition: \(np\) and \(n(1-p)\) should both be greater than 5.For \(p = 0.10\), \(n = 85\):\[ np = 85 \times 0.10 = 8.5, \quad n(1-p) = 85 \times 0.90 = 76.5 \]Both conditions are satisfied. Now, calculate the sample proportion:\[ \hat{p} = \frac{10}{85} \approx 0.1176 \]
03

Calculate Test Statistic

The test statistic for a proportion is calculated using the formula:\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Plug in the values:\[ Z = \frac{0.1176 - 0.10}{\sqrt{\frac{0.10 \times 0.90}{85}}} \approx 0.651 \]
04

Determine Critical Value and Make Decision

Use the standard normal distribution to find the critical value for \(\alpha = 0.05\): \(Z_{\alpha} = 1.645\).Since \(Z = 0.651 < Z_{\alpha} = 1.645\), we fail to reject the null hypothesis. There is not strong evidence that the proportion exceeds 0.10.
05

Calculate Type II Error Probability \(\beta\)

To find how likely it is that the test will not reject \(H_0\) if \(p = 0.15\), we need to calculate \(\beta\):Find the non-rejection point:\[ Z = \frac{\hat{p} - 0.10}{\sqrt{\frac{0.10 \times 0.90}{85}}} = 1.645 \]Thus, \(\hat{p} = 0.10 + 1.645 \times \sqrt{\frac{0.10 \times 0.90}{85}} \approx 0.138 \).Now, find \(\beta\) at \(p = 0.15\):\[ \beta = P(\hat{p} < 0.138 | p = 0.15) = P\left(\frac{\hat{p} - 0.15}{\sqrt{\frac{0.15 \times 0.85}{85}}} < \frac{0.138 - 0.15}{\sqrt{\frac{0.15 \times 0.85}{85}}}\right) \]Calculate the Z-value:\[ Z = \frac{0.138 - 0.15}{\sqrt{\frac{0.15 \times 0.85}{85}}} \approx -0.699 \]Look up this Z-value in the normal distribution: \(\beta \approx 0.242 \).
06

Determine Required Sample Size

To achieve a power of 0.90 (probability of correctly rejecting \(H_0\)), use the formula \(1-\beta = 0.90\). Set \(\beta = 0.10\) and solve:\[ Z_{1-\beta} = 1.282 \] (from Z-tables)For \(\alpha = 0.05 \), \(Z_{\alpha} = 1.645\).Rearrange the sample size formula:\[ n = \left(\frac{Z_{\alpha} \sqrt{p_0(1-p_0)} + Z_{1-\beta} \sqrt{p_1(1-p_1)}}{p_1 - p_0}\right)^2 \]Use \(p_0 = 0.10\) and \(p_1 = 0.15\):\[ n = \left(\frac{1.645 \sqrt{0.10 \times 0.90} + 1.282 \sqrt{0.15 \times 0.85}}{0.15 - 0.10}\right)^2 \approx 237.59 \]
07

Conclusion

Round up the calculated sample size to the nearest whole number.\[ n = 238 \]Thus, a sample size of 238 is required to have a power of 0.90 for \(p = 0.15\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Type I Error
A Type I Error occurs when we wrongly reject a true null hypothesis. It is also known as a 'false positive'. Imagine you're testing to see if a new drug is effective, and you conclude that it is, when in fact it isn't. This mistake can lead to incorrect actions based on bad evidence.

In hypothesis testing, the probability of making a Type I Error is denoted by \( \alpha \). An \( \alpha \) value of 0.05, for example, means there's a 5% risk of making a Type I Error. Therefore, the lower the \( \alpha \), the less likely we are to wrongly reject a true hypothesis.

In our crankshaft bearing study, if we choose \( \alpha = 0.05 \), it indicates that there's a 5% chance we might wrongly conclude that the proportion of bearings with excess roughness exceeds 0.10 when, in fact, it does not.
Type II Error
Type II Error, often referred to as a 'false negative', happens when we fail to reject a false null hypothesis. In layman's terms, it's when we mistakenly think there is no effect or difference when there truly is one.

The probability of a Type II Error is denoted by \( \beta \). Higher \( \beta \) values mean higher chances of making this error. Reducing \( \beta \) is crucial when we want to confidently identify real effects in experiments.

For instance, failing to identify that 15% of bearings exceed roughness due to an unusually low sample size or other factors leads to a Type II Error. In our study, \( \beta = 0.242 \) suggests there's a 24.2% chance of not detecting that more than 10% of bearings are rough, even if \( p = 0.15 \).
Sample Size Calculation
Determining the appropriate sample size is critical to achieving reliable results in hypothesis testing. Having too small a sample might lead to inaccurate conclusions, while too large a sample can be costly and time-consuming.

When calculating sample size, consider both Type I and Type II Error rates as well as the minimum detectable difference you wish to identify. The formula for sample size takes into account these factors along with expected proportions.

In our example with crankshaft bearings, to achieve a power of 0.90, which is the probability of correctly rejecting the null hypothesis, the sample size needed is 238. This ensures that if \( p = 0.15 \), there's a 90% chance of the test correctly identifying this, reducing the likelihood of a Type II Error to 10%.
Normal Approximation Condition
The normal approximation condition is key when dealing with sample proportions. It's used to ensure that the sample size is large enough to approximate the sampling distribution to a normal distribution.

This condition requires both \( np \) and \( n(1-p) \) to be greater than 5. This ensures the central limit theorem applies, thereby allowing us to use normal distribution assumptions in our calculations.

For the crankshaft bearing study, with \( n = 85 \) and \( p = 0.10 \), we calculated \( np = 8.5 \) and \( n(1-p) = 76.5 \), satisfying the normal approximation condition. This validation is crucial for the reliability of our Z-test results.
Test Statistic
In hypothesis testing, the test statistic is a standardized value that helps determine whether to reject the null hypothesis. It measures how far the sample statistic is from the hypothesized population parameter in terms of standard error.

The test statistic used for proportions is the Z-score, calculated as:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized proportion, and \( n \) is the sample size.

In our crankshaft bearing example, the Z-score came out to be approximately 0.651. This score is compared to a critical value from the standard normal distribution (1.645 for a 0.05 significance level). Since 0.651 is less than 1.645, we do not have enough evidence to reject the null hypothesis that only 10% of bearings exceed roughness standards.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An article in the British Medical Journal \(\left[{ }^{*}\right.\) Comparison of Treatment of Renal Calculi by Operative Surgery, Percutaneous Nephrolithotomy, and Extra-Corporeal Shock Wave Lithotrips" (1986, Vol. 292, pp. \(879-882\) ) ] repeated that percutaneous nephrolithotomy (PN) had a success rate in removing kidney stones of 289 of 350 patients. The traditional method was \(78 \%\) effective. (a) Is there evidence that the success rate for PN is greater than the historical success rate? Find the \(P\) -value. (b) Explain how the question in part (a) could be answered with a confidence interval.

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) \(z_{0}=0.4\)

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

A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees \(\mathrm{F}\), the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich") reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: 97.8,97.2,97.4,97.6 97.8,97.9,98.0,98.0,98.0,98.1,98.2,98.3,98.3,98.4,98.4 \(98.4,98.5,98.6,98.6,98.7,98.8,98.8,98.9,98.9,\) and \(99.0 .\) (a) Test the hypothesis \(H_{0}: \mu=98.6\) versus \(H_{1}: \mu \neq 98.6,\) using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that female body temperature is normally distributed. (c) Compute the power of the test if the true mean female body temperature is as low as \(98.0 .\) (d) What sample size would be required to detect a true mean female body temperature as low as 98.2 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 female body temperature.

An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with \(\sigma=60\) psi. A random sample of 12 specimens has a mean tensile strength of \(\bar{x}=3450\) psi. (a) Test the hypothesis that mean strength is 3500 psi. Use \(\alpha=0.01\) (b) What is the smallest level of significance at which you would be willing to reject the null hypothesis? (c) What is the \(\beta\) -error for the test in part (a) if the true mean is \(3470 ?\) (d) Suppose that you wanted to reject the null hypothesis with probability at least 0.8 if mean strength \(\mu=3470 .\) What sample size should be used? (e) Explain how you could answer the question in part (a) with a two-sided confidence interval on mean tensile strength.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.