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Heat through the glass 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: 1.11 1.07 1.11 1.07 1.12 1.08 1.08 1.18 1.18 1.18 1.12 Is there convincing evidence that the conductivity of this type of glass is greater than 1? Carry out a test to help you answer this question.

Step by step solution

01

State the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis \(H_0\) is that the mean conductivity is equal to 1, while the alternative hypothesis \(H_a\) is that the mean conductivity is greater than 1. Mathematically, \(H_0: \mu = 1\) and \(H_a: \mu > 1\).

02

Gather the Data

Identify the sample data: \(1.11, 1.07, 1.11, 1.07, 1.12, 1.08, 1.08, 1.18, 1.18, 1.18, 1.12\). This data represents the heat conductivity of the 11 randomly selected pieces of glass.

03

Calculate Sample Mean and Standard Deviation

Calculate the sample mean \(\bar{x}\) and standard deviation \(s\). The sample mean is \(\bar{x} = \frac{1.11 + 1.07 + 1.11 + 1.07 + 1.12 + 1.08 + 1.08 + 1.18 + 1.18 + 1.18 + 1.12}{11} \), which equals 1.117. The standard deviation \(s\) is calculated using the formula \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \), where \(n\) is the sample size (11).

04

Conduct the t-Test

To test the hypothesis, use a t-test for a single mean. Calculate the test statistic \( t \) using the formula \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \(\mu = 1\). Find \(t\) from the values obtained in Step 3 and compare \(t\) with the critical value from the t-distribution table at a chosen significance level (e.g., \(\alpha = 0.05\)).

05

Interpret the Results

Using the calculated \(t\) value and comparing it with the critical t value for 10 degrees of freedom (\( n - 1 = 11 - 1 = 10 \)), determine whether to reject or fail to reject the null hypothesis. If the calculated \(t\) is greater than the critical value, reject \(H_0\) in favor of \(H_a\).

06

State the Conclusion

Based on the comparison, if the test statistic \( t \) exceeds the critical value, conclude that there is convincing evidence that the mean conductivity of this type of glass is greater than 1.

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Key Concepts

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

t-test

The t-test is a statistical test that is used to compare the means of a sample with the theoretical mean from a population. In this context, it helps determine if our sample of glass pieces has a conductivity significantly greater than the baseline value of 1.
To perform a t-test, we calculate a test statistic based on the sample data. This statistic is then compared against a critical value from the t-distribution table.
Key aspects of a t-test include:

• The formula: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)
• \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesis mean, \( s \) is the standard deviation, and \( n \) is the sample size.
• The test requires the assumption of a normal distribution for the sample or it should not significantly deviate from normality.

The t-test helps make decisions based on statistically derived evidence, allowing us to infer if the mean of the sample significantly deviates from the expected result.

sample mean

The sample mean, represented as \( \bar{x} \), is the average value of a set of numbers taken from a population. It is one of the most commonly used measures of central tendency.
In the context of testing glass conductivity, calculating the sample mean provides an average value of the heat conductivity for our 11 glass pieces. This is crucial for determining how our sample compares to the hypothesized value of 1.
To calculate the sample mean:

• Add all of the observed conductivity values together: 1.11, 1.07, 1.11, 1.07, 1.12, 1.08, 1.08, 1.18, 1.18, 1.18, and 1.12.
• Divide this sum by the number of observations (11).
• Result: \( \bar{x} = 1.117 \).

The sample mean provides a pivotal value that serves as the foundation for the t-test calculation, offering insights into how similar or different the sample might be from the hypothesized average.

standard deviation

Standard deviation is a measure that indicates how much variation or dispersion exists from the average (mean) of a data set. It is symbolized as \( s \) in formulas.
In this scenario, standard deviation helps quantify the amount of variability in heat conductivities of the glass pieces. A smaller standard deviation indicates that the values tend to be close to the sample mean, whereas a larger standard deviation shows more spread out data.
To calculate the standard deviation:

• Determine each value's deviation from the sample mean (\( x_i - \bar{x} \)).
• Square these deviations.
• Sum all of these squared deviations.
• Divide by \( n-1 \), where \( n \) is the number of samples.
• Take the square root of this quotient.

This calculation results in the standard deviation value used in the t-test to understand and convey the reliability and precision of the sample mean.

null hypothesis

The null hypothesis, denoted as \( H_0 \), is a fundamental concept in hypothesis testing. It acts as the default assumption or starting point in any statistical analysis.
In this problem, the null hypothesis states that the mean conductivity of our glass samples is equal to 1, symbolically represented as \( H_0: \mu = 1 \).
Key points about the null hypothesis:

• It posits no effect or no difference, working as a starting assumption for statistical testing.
• The test aims to gather evidence to support or refute this assumption.
• We either 'reject' or 'fail to reject' the null hypothesis based on our statistical test results.

The null hypothesis plays a critical role because it formulates the basis for comparison against the alternative hypothesis (\( H_a \)), which is what we aim to prove or support based on our data – in this case, showing that the glass's conductivity might be greater than 1.