/*! 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 25 The desired percentage of \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The desired percentage of \(\mathrm{SiO}_{2}\) in a certain type of aluminous cement is \(5.5\). To test whether the true average percentage is \(5.5\) for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of \(\mathrm{SiO}_{2}\) in a sample is normally distributed with \(\sigma=.3\) and that \(\bar{x}=5.25\). a. Does this indicate conclusively that the true average percentage differs from \(5.5\) ? Carry out the analysis using the sequence of steps suggested in the text. b. If the true average percentage is \(\mu=5.6\) and a level \(\alpha=\) \(.01\) test based on \(n=16\) is used, what is the probability of detecting this departure from \(H_{0}\) ? c. What value of \(n\) is required to satisfy \(\alpha=.01\) and \(\beta(5.6)=.01 ?\)

Short Answer

Expert verified
a) Yes, the true average differs. b) Power is 85.08%. c) Sample size needed is 217.

Step by step solution

01

State the Hypotheses

To determine if the average \(\mathrm{SiO}_{2}\) percentage is different from 5.5, we set up the hypotheses:- Null Hypothesis \(H_0: \mu = 5.5\) (The true average percentage of \(\mathrm{SiO}_{2}\) is 5.5.)- Alternative Hypothesis \(H_a: \mu eq 5.5\) (The true average percentage of \(\mathrm{SiO}_{2}\) is not 5.5.)
02

Calculate the Test Statistic

The test statistic for the sample mean is calculated using the formula:\[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \]Substitute in the given values: \( \bar{x} = 5.25\), \( \mu_0 = 5.5\), \( \sigma = 0.3\), and \( n = 16 \):\[ z = \frac{5.25 - 5.5}{0.3 / \sqrt{16}} = \frac{-0.25}{0.075} = -3.33 \]
03

Determine the Critical Value(s)

For a two-tailed test at the \(\alpha = 0.01\) level, the critical z-values are \( z = \pm 2.576 \). These are derived from the standard normal distribution.
04

Compare Test Statistic and Critical Values

The calculated test statistic is \(-3.33\), which is less than \(-2.576\). This falls in the rejection region, meaning we reject the null hypothesis \( H_0 \).
05

Interpret the Results

Since \(-3.33\) is beyond the critical value of \(-2.576\), there is sufficient evidence to conclude that the true average percentage of \(\mathrm{SiO}_{2}\) is different from \(5.5\).
06

Calculate the Power for \( \mu = 5.6 \)

Power is the probability of rejecting \(H_0\) when \( \mu = 5.6 \). The critical values correspond to sample means of 5.322 and 5.678 (calculated from z = 2.576).The z-score for mean \(5.678\) under \(\mu = 5.6\):\[ z = \frac{5.678 - 5.6}{0.3/\sqrt{16}} = 1.04 \]The probability of z being greater than 1.04 is 0.1492, then power is \(1 - 0.1492 = 0.8508\).
07

Find the Required Sample Size for Specific \( \alpha \) and \( \beta \)

Solve for \(n\) using the formula:\[ z_{\alpha/2} + z_{\beta} = \frac{(\mu - \mu_0) \sqrt{n}}{\sigma} \]Using \(z_{\alpha/2} = 2.576\) and \(z_{\beta} = 2.326\) for \(\beta = 0.01\):\[ 2.576 + 2.326 = \frac{(5.6 - 5.5)\sqrt{n}}{0.3} \]\[ 4.902 = \frac{0.1 \sqrt{n}}{0.3} \]\[ \sqrt{n} = \frac{4.902 \times 0.3}{0.1} = 14.706 \]Thus, \(n = (14.706)^2 \approx 216.19\). Round up to 217 for suitable sample size.

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.

Statistical Inference
Statistical inference is a powerful tool that helps researchers draw conclusions about an entire population based on a sample. It involves a process of evaluating evidence from data to support or reject a hypothesis. In our case, the question is whether the average percentage of \( \mathrm{SiO}_{2} \) in aluminous cement is different from the known value of 5.5. By inspecting a sample's data, we can infer whether the production facility is operating within the expected norms or if deviations exist.

Hypothesis testing is a core part of statistical inference. It involves creating a null hypothesis \( H_0 \), reflecting the status quo (e.g., \( \mu = 5.5 \)) and an alternative hypothesis \( H_a \), indicating a deviation (e.g., \( \mu eq 5.5 \)). By comparing the real sample data to these hypotheses using statistical methods, like the z-test, we make calculated decisions about whether to accept or reject \( H_0 \).

The calculated z-score (a measure of standard deviations away from the mean of the distribution) helps determine if our sample mean significantly deviates from what we expected. In our exercise, a z-score of \(-3.33\) suggests a significant deviation since it falls beyond the critical z-value of \(-2.576\), leading to the conclusion that there is a statistically significant difference.
Significance Level
The significance level, often denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by researchers to assess the credibility of the statistical evidence against \( H_0 \).

In practice, common significance levels are 0.05, 0.01, or 0.001. A lower \( \alpha \) means stricter criteria to declare a result statistically significant. In the exercise, \( \alpha \) is set to 0.01, which implies that we accept only a 1% chance of mistakenly rejecting \( H_0 \).

Choosing the right \( \alpha \) depends on context. For critical fields like medicine, a low \( \alpha \) minimizes false positives, but in exploratory research, a higher \( \alpha \) might be acceptable to flag potential leads. The selected \( \alpha \) influences the critical values of the test statistic, which in turn affect the decision about whether sample data provides sufficient evidence to reject \( H_0 \).
Sample Size Determination
Deciding on the sample size is an important aspect of planning a study. The sample size must be sufficient to ensure reliable results while considering resources like time and money. In hypothesis testing, the sample size can affect the power of the test, which is the probability of correctly rejecting \( H_0 \) when it is false.

In our exercise, we determine the sample size needed to achieve desired sensitivity and specificity in results. The formula involves parameters such as \( \alpha \), \( \beta \) (probability of a Type II error), the standard deviation \( \sigma \), and the desired effect size (difference between \( \mu \) and \( \mu_0 \)).

The calculated sample size in the step-by-step solution is 217. This is derived by correlating the given \( \alpha = 0.01 \) and \( \beta = 0.01 \) with \( \mu_0 = 5.5 \) and \( \mu = 5.6 \), ensuring that both the statistical significance and power are met effectively. Increasing the sample size reduces standard errors and hence narrows confidence intervals, leading to more precise estimates of the population parameters.

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

The article "Orchard Floor Management Utilizing SoilApplied Coal Dust for Frost Protection" (Agri. and Forest Meteorology, 1988: 71-82) reports the following values for soil heat flux of eight plots covered with coal dust. \(\begin{array}{llllllll}34.7 & 35.4 & 34.7 & 37.7 & 32.5 & 28.0 & 18.4 & 24.9\end{array}\) The mean soil heat flux for plots covered only with grass is 29.0. Assuming that the heat-flux distribution is approximately normal, does the data suggest that the coal dust is effective in increasing the mean heat flux over that for grass? Test the appropriate hypotheses using \(\alpha=.05\).

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using MINITAB, resulting in the accompanying output. $$ \begin{array}{rrrrrr} \text { Variable } & \text { M Mean StDev } & \text { SEMean } & \text { ZP- Value } \\ \text { lifetime } 50 & 738.44 & 38.20 & 5.40-2.14 & 0.016 \end{array} $$ What conclusion would be appropriate for a significance level of \(.05\) ? A significance level of \(.01\) ? What significance level and conclusion would you recommend?

A university library ordinarily has a complete shelf inventory done once every year. Because of new shelving rules instituted the previous year, the head librarian believes it may be possible to save money by postponing the inventory. The librarian decides to select at random 1000 books from the library's collection and have them searched in a preliminary manner. If evidence indicates strongly that the true proportion of misshelved or unlocatable books is less than \(.02\), then the inventory will be postponed. a. Among the 1000 books searched, 15 were misshelved or unlocatable. Test the relevant hypotheses and advise the librarian what to do (use \(\alpha=.05\) ). b. If the true proportion of misshelved and lost books is actually .01, what is the probability that the inventory will be (unnecessarily) taken? c. If the true proportion is \(.05\), what is the probability that the inventory will be postponed?

After a period of apprenticeship, an organization gives an exam that must be passed to be eligible for membership. Let \(p=P\) (randomly chosen apprentice passes). The organization wishes an exam that most but not all should be able to pass, so it decides that \(p=.90\) is desirable. For a particular exam, the relevant hypotheses are \(H_{0}: p=.90\) versus the alternative \(H_{\mathrm{a}}: p \neq 90\). Suppose ten people take the exam, and let \(X=\) the number who pass. a. Does the lower-tailed region \(\\{0,1, \ldots, 5\\}\) specify a level \(.01\) test? b. Show that even though \(H_{\mathrm{a}}\) is two-sided, no two-tailed test is a level \(.01\) test. c. Sketch a graph of \(\beta\left(p^{\prime}\right)\) as a function of \(p^{\prime}\) for this test. Is this desirable?

Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. The article "A Field Trial of Ketamine Anesthesia in the Horse" (Equine Vet. J., 1984: 176-179) reports that for a sample of \(n=73\) horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was \(18.86 \mathrm{~min}\) and the standard deviation was \(8.6 \mathrm{~min}\). Does this data suggest that true average lateral recumbency time under these conditions is less than \(20 \mathrm{~min}\) ? Test the appropriate hypotheses at level of significance . 10 .

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.