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Chapter 9: Problem 48

Newly purchased tires of a certain type are supposed to be filled to a pressure of \(30 \mathrm{lb} / \mathrm{in}^{2}\). Let \(\mu\) denote the true average pressure. Find the \(P\)-value associated with each given \(z\) statistic value for testing \(H_{0}: \mu=30\) versus \(H_{\mathrm{a}}: \mu \neq 30\). a. \(2.10\) b. \(-1.75\) c. \(-.55\) d. \(1.41\) e. \(-5.3\)

Short Answer

Expert verified
a. 0.0358, b. 0.0802, c. 0.5824, d. 0.1586, e. 鈮�0

Step by step solution

01

Understand the Null and Alternative Hypotheses

The null hypothesis ( H_0) is that the true average pressure is 30 lb/in虏. The alternative hypothesis ( H_a) is that the true average pressure is not equal to 30 lb/in虏. We are performing a two-tailed test.
02

Identify the Given z Statistic Values

The provided z statistic values are: \( z = 2.10 \), \( z = -1.75 \), \( z = -0.55 \), \( z = 1.41 \), and \( z = -5.3 \).
03

Determine the P-value for z = 2.10

Since we are performing a two-tailed test, we need to find the two-tailed P-value associated with \( z = 2.10 \). Using the standard normal distribution table or a calculator, find \( P(Z > 2.10) = 0.0179 \), so the two-tailed P-value is \( 2 \times 0.0179 = 0.0358 \).
04

Determine the P-value for z = -1.75

Similarly, for \( z = -1.75 \), find \( P(Z < -1.75) = 0.0401 \). Thus, the two-tailed P-value is \( 2 \times 0.0401 = 0.0802 \).
05

Determine the P-value for z = -0.55

For \( z = -0.55 \), find \( P(Z < -0.55) = 0.2912 \). Thus, the two-tailed P-value is \( 2 \times 0.2912 = 0.5824 \).
06

Determine the P-value for z = 1.41

For \( z = 1.41 \), find \( P(Z > 1.41) = 0.0793 \). Thus, the two-tailed P-value is \( 2 \times 0.0793 = 0.1586 \).
07

Determine the P-value for z = -5.3

For \( z = -5.3 \), find \( P(Z < -5.3) \), which is very small, approximately 0. So the two-tailed P-value is essentially 0.

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Key Concepts

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

Z-test
A Z-test is a type of statistical test that helps determine if there is a significant difference between sample and population means. It's particularly useful when dealing with large sample sizes or when the variance is known.
For our exercise, the Z-test is used to decide if the average tire pressure deviates from the specified value of 30 lb/in虏. Here's how it generally works:
  • The population variance is assumed to be known.
  • The sample is believed to follow a normal distribution.
  • We compare the Z statistic against the standard normal distribution to make conclusions.
In our context, once we calculate the Z-statistic, we take it a step further by finding the P-value. This value tells us the probability of observing a result as extreme as the test statistic. With the Z-test, we can objectively assess the likelihood of deviations under the null hypothesis.
P-value
The P-value is an essential concept in hypothesis testing, indicating how extreme the observed data is under the null hypothesis. It helps determine the strength of the results.
  • A low P-value (typically 鈮� 0.05) suggests strong evidence against the null hypothesis, implying it's unlikely to occur by random chance.
  • A high P-value (greater than 0.05) suggests weak evidence against the null hypothesis.
In our exercise, each Z-value was converted to a P-value using statistical tables or calculators, reflecting the probability that the measured outcome occurs under the assumption the null hypothesis is true. This provides clarity on whether the actual average tire pressure significantly deviates from the expected 30 lb/in虏. Although our context uses a standard 5% significance level, it's crucial to understand each situation might have different threshold levels for what is deemed significant.
Normal Distribution
The normal distribution is a fundamental concept in statistics often referred to as a bell curve due to its shape. This distribution is valuable in hypothesis testing because many statistical tests
  • assume the data follows a normal distribution, especially with large sample sizes,
  • are based on the properties of this bell-shaped curve, focusing on the mean and standard deviation.
In our scenario, we assume the tire pressure data follows a normal distribution. This inference allows us to use a Z-test to analyze the data. The mean of the distribution is our hypothesis value (30 lb/in虏), and the standard deviation is the spread of data points around this mean. Such assumptions are crucial when interpreting results from statistical tests like the Z-test because they ensure the accuracy of calculated P-values and subsequent conclusions.
Two-tailed test
A two-tailed test is a type of hypothesis test that checks for deviations in both directions away from the hypothesized value. This test is used when we want to know if the sample mean is either significantly lower or higher than the hypothesized population mean.
In the context of the tire exercise, we use a two-tailed test because:
  • We're interested in deviations on both sides of 30 lb/in虏.
  • We aim to determine if tire pressures are unexpectedly high or low.
Mathematically, it involves considering both tails of the normal distribution curve. Therefore, the P-value must be doubled to factor in both possibilities. This approach ensures we capture any significant deviation from the null hypothesis, regardless of the direction. Understanding two-tailed tests is crucial since they provide a comprehensive view of potential variations from expected results.

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Most popular questions from this chapter

The incidence of a certain type of chromosome defect in the U.S. adult male population is believed to be 1 in 75. A random sample of 800 individuals in U.S. penal institutions reveals 16 who have such defects. Can it be concluded that the incidence rate of this defect among prisoners differs from the presumed rate for the entire adult male population? a. State and test the relevant hypotheses using \(\alpha=.05\). What type of error might you have made in reaching a conclusion? b. What \(P\)-value is associated with this test? Based on this \(P\)-value, could \(H_{0}\) be rejected at significance level .20?

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge-water temperature above \(150^{\circ}, 50\) water samples will be taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

The amount of shaft wear (.0001 in.) after a fixed mileage was determined for each of \(n=8\) internal combustion engines having copper lead as a bearing material, resulting in \(\bar{x}=3.72\) and \(s=1.25\). a. Assuming that the distribution of shaft wear is normal with mean \(\mu\), use the \(t\) test at level . 05 to test \(H_{0}: \mu=3.50\) versus \(H_{\mathrm{a}}: \mu>3.50\). b. Using \(\sigma=1.25\), what is the type II error probability \(\beta\left(\mu^{\prime}\right)\) of the test for the alternative \(\mu^{\prime}=4.00\) ?

Let the test statistic \(Z\) have a standard normal distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), rejection region \(z \geq 1.88\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}\), rejection region \(z \leq-2.75\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), rejection region \(z \geq 2.88\) or \(z \leq\) \(-2.88\)

A hot-tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 32 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(17.5 \mathrm{~min}\) and \(2.2 \mathrm{~min}\), respectively. Does this data cast doubt on the company's claim? Compute the \(P\)-value and use it to reach a conclusion at level \(.05\) (assume that the heating-time distribution is approximately normal).
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