/*! 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 19 The melting point of each of 16 ... [FREE SOLUTION] | 91Ó°ÊÓ

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The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\bar{x}=94.32\). Assume that the distribution of the melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{\mathrm{a}}: \mu \neq 95\) using a two- tailed level \(.01\) test. b. If a level \(.01\) test is used, what is \(\beta(94)\), the probability of a type II error when \(\mu=94\) ? c. What value of \(n\) is necessary to ensure that \(\beta(94)=.1\) when \(\alpha=.01\) ?

Short Answer

Expert verified
a. Do not reject \(H_0\); b. \(\beta(94) \approx 0.4325\); c. \(n=55\) is necessary.

Step by step solution

01

Determine the Test Statistic for Part (a)

Given: - Sample mean \(\bar{x} = 94.32\)- Population standard deviation \(\sigma = 1.20\)- Sample size \(n = 16\)- Null hypothesis \(H_{0}: \mu = 95\)The test statistic \(z\) is calculated using the formula:\[z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]Plug in the values:\[z = \frac{94.32 - 95}{1.20 / \sqrt{16}} = \frac{-0.68}{0.3} = -2.27\]
02

Determine Critical Z-Scores for Part (a)

Since we are conducting a two-tailed test at level \(\alpha = 0.01\), we split the \(\alpha\) into two tails, each with \(0.005\).The critical z-scores for a two-tailed test at \(\alpha = 0.01\) are approximately \(-2.576\) and \(2.576\).
03

Test the Hypothesis for Part (a)

Compare the calculated test statistic \(z = -2.27\) with the critical values:- Since \(-2.27\) is not less than \(-2.576\) and not greater than \(2.576\), it falls within the region of acceptance.Thus, we do not reject the null hypothesis \(H_0\: \mu = 95\).
04

Determine Beta for Part (b)

\( \beta(94) \) is the probability of a Type II error given \( \mu = 94 \) when the null hypothesis is that \( \mu = 95 \).Calculate the non-central z-score for \(\mu = 94\):\[z = \frac{94 - 95}{1.20 / \sqrt{16}} = \frac{-1}{0.3} = -3.33\]This corresponding power curve (using mean \(\mu = 94\)) and the critical z-scores \(-2.576\) and \(2.576\) results in an integral for the probability of not rejecting \(\mu = 94\). Given standard tables or calculators, \(\beta(94)\) is approximately 0.4325.
05

Calculate Sample Size for Part (c)

We need \(n\) such that \(\beta(94) = 0.1\) when \(\alpha = 0.01\). Use the relation:The equation relating power to sample size is:\[\beta = \bar{Z}(Z_{\alpha/2} - (\mu_1 - \mu_0)/\sigma_{\bar{X}})\]Where:- \(Z_{\alpha/2} = 2.576\) for \(\alpha = 0.01\)- Determine \(Z_{1-\beta}\) for \(\beta = 0.1\), resulting in \(Z_{0.9}= 1.28\)- Solve the inequality for \(n\):\[2.576 - \frac{1 \cdot \sqrt{n}}{1.2} = 1.28\]\[\sqrt{n} = \frac{2.576 - 1.28}{0.833}\]\[ n = \lceil(1.448 \times 0.833)^{-2}\rceil = 54.21 \approx 55 \]
06

Solution Summary

For part (a), we do not reject the null hypothesis. For part (b), the probability of a type II error \(\beta(94)\) is approximately 0.4325. To ensure \(\beta(94) = 0.1\) when \(\alpha = 0.01\), \(n = 55\) is necessary for part (c).

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Key Concepts

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

Type I and Type II Errors
In hypothesis testing, understanding Type I and Type II errors is crucial. These errors represent mistakes we can make when deciding whether to reject or accept a hypothesis.

A **Type I Error** occurs when we reject a true null hypothesis. In simpler terms, it's a false positive. We claim there is an effect when there isn't one. This error is controlled by setting a significance level, denoted by \( \alpha \), which is the probability of making a Type I error. In our example, this level is 0.01, meaning we are willing to accept a 1% chance of incorrectly rejecting the null hypothesis.

On the other hand, a **Type II Error** happens when we fail to reject a false null hypothesis. This is like a false negative, where we claim there is no effect when there truly is one. The probability of making a Type II error is denoted by \( \beta \). In problems like the example given, part of the task is to find \( \beta \) for certain parameters, such as when \( \mu = 94 \), where it was calculated to be approximately 0.4325.

Balancing the probabilities of these errors is essential in any test design. Lowering one type of error often increases the other, making it important to choose \( \alpha \) and \( \beta \) carefully based on the context of the test and desired outcomes.
Normal Distribution
The term "normal distribution" refers to the bell-shaped curve that describes how values of a variable are distributed. Many phenomena in natural and social sciences follow this pattern.

In our example problem, we assume that the distribution of the melting point is normal with a known standard deviation, \( \sigma = 1.20 \). What makes the normal distribution special is that it is completely characterized by its mean (\( \mu \)) and standard deviation (\( \sigma \)).

The properties of a normal distribution include:
  • The curve is symmetric around the mean.
  • Approximately 68% of the data lies within one standard deviation of the mean, about 95% within two, and 99.7% within three.
  • The total area under the curve is 1, which represents the entirety of the event probability.
Recognizing that data follows a normal distribution allows us to employ z-scores in hypothesis testing, giving a standardized way to compare scores from different distributions by transforming them based on the mean and standard deviation.

This approach was key in the step-by-step solution. We calculated a test statistic \( z \) using the standard formula for normal distributions to determine whether to accept or reject the hypothesis.
Sample Size Calculation
Ensuring the correct sample size is crucial when planning experiments or studies because it affects the reliability and accuracy of results.

To determine the required sample size, we need to consider the desired levels of \( \alpha \) and \( \beta \). In the given problem, we wanted \( \beta = 0.1 \) with \( \alpha = 0.01 \). Here, calculating the correct sample size involves using the relationship between z-scores and the desired probabilities.

The usual formula to calculate sample size is based on making sure the test has enough power (which is 1-\( \beta \)) to detect a true effect. The exact formula involves:
  • \( Z_{\alpha/2} \), which is the z-score that corresponds to half of the \( \alpha \) level, representing the cutoff for statistical significance in a two-tailed test.
  • \( Z_{1-\beta} \), the z-score corresponding to the test's power, indicating the probability of correctly rejecting a false null hypothesis.
By rearranging the formula within the context of the specific test parameters, it was found that a sample size of 55 is needed to achieve the desired error probabilities for the given test of the melting points.

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Most popular questions from this chapter

A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood. Does this suggest that the actual percentage of type A donations differs from \(40 \%\), the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of \(.01\). Would your conclusion have been different if a significance level of \(.05\) had been used?

For a fixed alternative value \(\mu^{\prime}\), show that \(\beta\left(\mu^{\prime}\right) \rightarrow 0\) as \(n \rightarrow \infty\) for either a one-tailed or a two-tailed \(z\) test in the case of a normal population distribution with known \(\sigma\).

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): a. \(H_{0}: \mu=100, H_{\mathrm{a}}: \mu>100\) b. \(H_{0}: \sigma=20, H_{\mathrm{a}}: \sigma \leq 20\) c. \(H_{0}: p \neq .25, H_{\mathrm{a}}: p=.25\) d. \(H_{0}: \mu_{1}-\mu_{2}=25, H_{\mathrm{a}}: \mu_{1}-\mu_{2}>100\) e. \(H_{0}: S_{1}^{2}=S_{2}^{2}, H_{\mathrm{a}}: S_{1}^{2} \neq S_{2}^{2}\) f. \(H_{0}: \mu=120, H_{\mathrm{a}}: \mu=150\) g. \(H_{0}: \sigma_{1} / \sigma_{2}=1, H_{\mathrm{a}}: \sigma_{1} / \sigma_{2} \neq 1\) h. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30, respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)? e. For a level \(.05\) test, what conclusion would you reach?

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.

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