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An article in the ASCE Journal of Energy Engineering (1999, Vol. \(125,\) pp. \(59-75\) ) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures \(\left({ }^{\circ} \mathrm{C}\right)\) reported were as follows: \(23.01,22.22,22.04,22.62,\) and \(22.59 .\) a. Test the hypotheses \(H_{0}: \mu=22.5\) versus \(H_{1}: \mu \neq 22.5\), using \(\alpha=0.05 .\) Find the \(P\) -value. b. Check the assumption that interior temperature is normally distributed. c. Compute the power of the test if the true mean interior temperature is as high as \(22.75 .\) d. What sample size would be required to detect a true mean interior temperature as high as 22.75 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 interior temperature.

Short Answer

Expert verified
The hypothesis is not rejected, data is normally distributed, power is lower than 0.9, larger sample size is needed, CI includes the hypothesized mean.

Step by step solution

01

Calculate the Sample Mean and Standard Deviation

First, find the sample mean (\( \overline{x} \)) and the sample standard deviation (\( s \)). Given the data: 23.01, 22.22, 22.04, 22.62, 22.59.\[ \overline{x} = \frac{23.01 + 22.22 + 22.04 + 22.62 + 22.59}{5} = 22.496 \]The sample standard deviation \( s \) is calculated by:\[ s = \sqrt{\frac{\sum(x_i - \overline{x})^2}{n-1}} \]where \( x_i \) are individual observations and \( n = 5 \) is the sample size.
02

Perform the Hypothesis Test

Use the sample mean and standard deviation to perform a two-tailed t-test.The hypotheses are:\[ H_0: \mu = 22.5 \text{ vs. } H_1: \mu eq 22.5 \]Calculate the t-statistic:\[ t = \frac{\overline{x} - \mu}{s / \sqrt{n}} \]Check the degrees of freedom \( df = n - 1 = 4 \). Compare the t-statistic to the critical value from the t-table for \( \alpha = 0.05 \). Calculate the p-value using the t-distribution.
03

Assess Normality Assumption

Use the Shapiro-Wilk test, which is designed to test the normality of a sample. Alternatively, you can construct a Q-Q plot or histogram. If the p-value from the Shapiro-Wilk test is greater than 0.05, we fail to reject the null hypothesis that the data is normally distributed.
04

Calculate the Power of the Test

To find the power, first calculate the non-centrality parameter:\[ \delta = \frac{\text{difference in means}}{s / \sqrt{n}} = \frac{22.75 - 22.5}{s / \sqrt{5}} \]The power of the test is then calculated using this non-centrality parameter against the t-distribution with \( df = 4 \) for a true mean of 22.75.
05

Determine Required Sample Size for Desired Power

To find the sample size \( n \) for a desired power of 0.9, use the relationship:\[ z_{\alpha/2} + z_{power} = \frac{\delta}{s / \sqrt{n}} \]where \( z_{\alpha/2} \) and \( z_{power} \) are the z-values for α/2 and power 0.9 respectively. Solve this equation for \( n \).
06

Construct a Confidence Interval

Construct a 95% confidence interval for the mean:\[ \overline{x} \pm t_{\alpha/2} \times \left( \frac{s}{\sqrt{n}} \right) \]Compare the bounds of the confidence interval with \( \mu = 22.5 \). If \( 22.5 \) falls outside this interval, reject \( H_0 \).

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Key Concepts

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

t-test
When you have a small sample size and want to compare the sample mean to a known value, a t-test helps you determine if there's a significant difference or not. In this exercise, the null hypothesis (H_0: \mu = 22.5) tests if the average interior temperature is 22.5°C. The alternative hypothesis (H_1: \mu eq 22.5) suggests it's different. To assess these hypotheses, we calculate the t-statistic. It's done using the formula: \[ t = \frac{\overline{x} - \mu}{s / \sqrt{n}} \] where \( \overline{x} \) is the sample mean, \( \mu \) is the hypothesized mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Check the calculated t-value against the critical t-value from statistical tables at your chosen significance level (\( \alpha \)). Here, \( \alpha = 0.05 \), meaning you're allowing a 5% chance of rejecting the null hypothesis when it's actually true (type I error). If the t-value is beyond this critical range, it indicates a statistically significant difference, leading you to reject \( H_0 \).

Additionally, the p-value shows the probability of observing the test results under the null hypothesis. A p-value less than \( \alpha \) indicates strong evidence against \( H_0 \).
normality assumption
Before performing a t-test, it's crucial to ensure your data is approximately normally distributed. This is because the t-test relies on the normality assumption for small sample sizes (typically \( n < 30 \)).

The Shapiro-Wilk test is a common tool to assess normality. It provides a p-value to gauge whether your data deviates from a normal distribution. If this p-value is greater than 0.05, you don't have enough evidence to say your data isn't normal, thus, you can proceed with the t-test safely.

Alternatively, graphical methods like Q-Q plots or histograms also help visualize data normality. A Q-Q plot compares your data quantiles against a normal distribution. If points lie along the line, your data is likely normal. Histograms offer a visual check – a bell-shaped curve suggests normality.

The normality check is essential because if your data is far from normal, your t-test results might not be valid, leading to incorrect conclusions.
confidence interval
To further validate hypothesis testing, constructing a confidence interval (CI) offers another perspective. A 95% CI gives a range within which we expect the true mean to fall 95% of the time. In our example, the formula is: \[ \overline{x} \pm t_{\alpha/2} \times \left( \frac{s}{\sqrt{n}} \right) \] where \( t_{\alpha/2} \) is the t-value for \( 1 - \alpha/2 \) confidence, \( \overline{x} \) is the sample mean, \( s \) is the standard deviation, and \( n \) is the sample size.

If the hypothesized mean (22.5) lies outside this interval, it provides evidence against the null hypothesis, aligning with p-value results. Thus, CIs offer a confidence range rather than a simple yes/no hypothesis test result, crowning it a valuable tool for deeper insight.

Also, wider CIs suggest more variability in data while narrower CIs imply more precise estimates of the mean. So, confidence intervals not only support decision-making in hypothesis tests but also reveal estimation reliability.
sample size calculation
Determining the needed sample size for sufficient statistical power is an important step. If you want the power of a test to be at least 0.9, it ensures a 90% probability of correctly rejecting a false null hypothesis (type II error).

To compute required sample size \( n \), use this equation: \[ z_{\alpha/2} + z_{power} = \frac{\delta}{s / \sqrt{n}} \] where \( z_{\alpha/2} \) and \( z_{power} \) are standard normal z-values corresponding to the desired significance level and power.

The non-centrality parameter \( \delta \) relates to the difference you'd like to detect (true mean - hypothesized mean). Solving for \( n \) gives you the minimum sample size needed, ensuring your test's reliability.

Appropriate sample size benefits testing accuracy, saving resources, and avoiding unnecessary sampling. It leads to more reliable and confident conclusions and prevents weak statistical power that could miss a true effect.

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Most popular questions from this chapter

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test \(H_{0}: \mu=175\) millimeters versus \(H_{1}: \mu>175\) millimeters, using the results of \(n=10\) samples. a. Find the type I error probability \(\alpha\) if the critical region is \(\bar{x}>185\) b. What is the probability of type II error if the true mean foam height is 185 millimeters? c. Find \(\beta\) for the true mean of 195 millimeters.

A hypothesis will be used to test that a population mean equals 5 against the alternative that the population mean is less than 5 with known variance \(\sigma .\) What is the critical value for the test statistic \(Z_{0}\) for the following significance levels? a. 0.01 b. 0.05 c. 0.10

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