/*! 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 74 How well materials conduct heat ... [FREE SOLUTION] | 91Ó°ÊÓ

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How well materials conduct heat matters when designing houses, for example. Conductivity is measured in terms of watts of heat power transmitted per square meter of surface per degree Celsius of temperature difference on the two sides of the material. In these units, glass has conductivity about \(1 .\) The National Institute of Standards and Technology provides exact data on properties of materials. Here are measurements of the heat conductivity of 11 randomly selected pieces of a particular type of glass: \({ }^{22}\) \(\begin{array}{llllllllllll}1.11 & 1.07 & 1.11 & 1.07 & 1.12 & 1.08 & 1.08 & 1.18 & 1.18 & 1.18 & 1.12\end{array}\) (a) Is there convincing evidence that the mean conductivity of this type of glass is greater than \(1 ?\) (b) Given your conclusion in part (a), which kind of mistake-a Type I error or a Type II error - could you have made? Explain what this mistake would mean in context.

Short Answer

Expert verified
There is evidence the mean conductivity is greater than 1. A Type I error could occur, meaning we wrongly conclude it is greater than 1.

Step by step solution

01

State Hypotheses

We need to determine whether the mean heat conductivity of this type of glass is greater than 1. Set up the null hypothesis (H_0) as \( \mu = 1 \) (mean conductivity is 1) and the alternative hypothesis (H_a) as \( \mu > 1 \) \( \) (mean conductivity is greater than 1).
02

Calculate Sample Mean and Standard Deviation

Compute the sample mean (\overline{x}) and sample standard deviation (s) of the given data. The data values are: 1.11, 1.07, 1.11, 1.07, 1.12, 1.08, 1.08, 1.18, 1.18, 1.18, 1.12. \[\overline{x} = \frac{1.11 + 1.07 + 1.11 + 1.07 + 1.12 + 1.08 + 1.08 + 1.18 + 1.18 + 1.18 + 1.12}{11} = 1.1182\]Calculate \( s \) using the standard formula for the sample standard deviation.
03

Determine Test Statistic

Use the sample mean and standard deviation to compute the test statistic using the formula for the t-statistic: \[\t = \frac{\overline{x} - \mu}{s/\sqrt{n}}\]where \(\mu = 1\), \(\overline{x} = 1.1182\), \(s\) is the sample standard deviation, and \(n = 11\).
04

Find Critical Value or P-value

Using a chosen significance level (commonly \(\alpha = 0.05\)), find the critical value or p-value corresponding to the computed t-statistic. Compare the t-statistic against critical t-values from a t-distribution table with \( n-1 = 10 \) degrees of freedom. Alternatively, compute the p-value for the one-tailed test to check the likelihood of observing a test statistic as extreme as, or more extreme than, the observed value.
05

Make a Decision

If the t-statistic is larger than the critical value or if the p-value is less than the significance level \(\alpha\), reject the null hypothesis. This means there is convincing evidence that the mean conductivity of the glass is greater than 1.
06

Error Analysis

Since we are testing if the mean is greater than 1 and we reject H_0 , we could potentially make a Type I error (rejecting a true null hypothesis). In this context, a Type I error means concluding that the mean conductivity is greater than 1 when it actually is not.

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Key Concepts

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

Conductivity
Conductivity refers to the ability of a material to conduct heat, which is crucial for applications like home insulation. It is quantified by measuring how much heat power, in watts, is transmitted through a material per square meter at a temperature difference of one degree Celsius. This property helps determine how efficient a material is at transferring heat. For instance, glass has a typical conductivity value of around 1 in these units.
Understanding conductivity is important because it affects energy efficiency in buildings. Materials with high conductivity allow more heat to pass through, which can be undesirable during temperature extremes. Conversely, materials with low conductivity help insulate buildings, keeping them cool in summer and warm in winter.
Therefore, assessing the true mean conductivity of materials like glass is crucial for designing energy-efficient buildings. This process often involves statistical hypothesis testing, where we use sample data to infer characteristics about the overall population of the material.
Type I Error
In statistical hypothesis testing, a Type I error occurs when we reject a true null hypothesis. This means that we incorrectly conclude that there is an effect or a difference when really there is none. In the context of evaluating the conductivity of glass, a Type I error would happen if we conclude that the mean conductivity is greater than 1 when, in fact, it is not.
Thinking about hypothesis testing, we set up hypotheses to see if sample data can support a specific claim about a population parameter (like mean conductivity). The null hypothesis usually states that there is no effect or difference, while the alternative hypothesis shows the opposite. If we reject the null hypothesis based on our sample, there's a risk of a Type I error.
This error can lead to poor decision-making. For example, if builders rely on the flawed conclusion that a certain type of glass is better insulator than it is, they may choose materials that do not actually yield the desired energy efficiency, leading to increased heating or cooling costs.
Sample Standard Deviation
Sample standard deviation provides an estimate of the variability or spread in a set of data points. It reflects how much the individual data values typically differ from the sample mean. For our glass conductivity data, it highlights how consistent the glass pieces are in conducting heat.
To calculate the sample standard deviation, use the formula:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \overline{x})^2} \]where \(s\) is the sample standard deviation, \(n\) is the number of observations, \(x_i\) are the data points, and \(\overline{x}\) is the sample mean.
Understanding the sample standard deviation is important for conducting hypothesis tests, like the t-test used here. It is crucial because it provides insight into the data's variability, which affects how we compute the test statistic. A smaller standard deviation indicates data points closer to the mean, while a larger one suggests more spread out data.
Critical Value
The critical value is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It is based on the chosen significance level, usually denoted by \(\alpha\), which represents the probability of making a Type I error.
In our case, when checking if glass's mean conductivity is greater than 1, we calculate a t-statistic. This involves our sample data's mean and standard deviation. We then compare this statistic to a critical value from the t-distribution table with \( n-1 \) degrees of freedom, where \( n \) is the sample size (11 in our case, so 10 degrees of freedom).
If the calculated t-statistic is greater than the critical value, we reject the null hypothesis, suggesting the mean conductivity is indeed greater than 1. Alternatively, if you determine a p-value (the probability of observing such an extreme test statistic by chance), you compare it to \(\alpha\). A p-value lower than \(\alpha\) indicates sufficient evidence to reject the null hypothesis. Thus, critical values and p-values guide our decisions in hypothesis testing by quantifying the evidence against the null hypothesis.

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Most popular questions from this chapter

Does listening to music while studying hinder students' learning? Two AP Statistics students designed an experiment to find out. They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference in the number of words recalled (music- silence) was recorded for each subject. A paired \(t\) test on the differences yielded \(t=-3.01\) and \(P\) -value \(=0.0027\) (a) State appropriate hypotheses for the paired \(t\) test. Be sure to define your parameter. (b) What are the degrees of freedom for the paired \(t\) test? (c) Interpret the \(P\) -value in context. What conclusion should the students draw? (d) Describe a Type I error and a Type II error in this setting. Which mistake could students have made based on your answer to part (c)?

You read that a statistical test at the \(\alpha=0.01\) level has probability 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative?

Explaining confidence (8.2) Here is an explanation from a newspaper concerning one of its opinion polls. Explain what is wrong with the following statement. For \(a\) poll of 1,600 adults, the variation due to sampling error is no more than three percentage points either way. The error margin is said to be valid at the 95 percent confidence level. This means that, if the same questions were repeated in 20 polls, the results of at least 19 surveys would be within three percentage points of the results of this survey.

Significance tests \(A\) test of \(H_{0}: p=0.5\) versus \(H_{a}: p>0.5\) has test statistic \(z=2.19\) (a) What conclusion would you draw at the \(5 \%\) significance level? At the \(1 \%\) level? (b) If the alternative hypothesis were \(H_{a}: p \neq 0.5,\) what conclusion would you draw at the \(5 \%\) significance level? At the \(1 \%\) level?

Fresh coffee People of taste are supposed to prefer fresh-brewed coffee to the instant variety. On the other hand, perhaps many coffee drinkers just want their caffeine fix. A skeptic claims that only half of all coffee drinkers prefer fresh-brewed coffee. To test this claim, we ask a random sample of 50 coffee drinkers in a small city to take part in a study. Each person tastes two unmarked cupsone containing instant coffee and one containing fresh-brewed coffee-and says which he or she prefers. We find that 36 of the 50 choose the fresh coffee. (a) Do these results give convincing evidence that coffee drinkers favor fresh-brewed over instant coffee? (b) We presented the two cups to each coffee drinker in a random order, so that some people tasted the fresh coffee first, while others drank the instant coffee first. Why do you think we did this?

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