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Ch. 7 - Hypothesis Testing with One Sample
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 7, Problem 7.2.11

Graphical Analysis In Exercises 9–12, match the P-value or z-statistic with the graph that represents the corresponding area. Explain your reasoning.


z = -2.37


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Step 1: Understand the z-statistic value provided in the problem. A z-statistic of -2.37 indicates that the value is 2.37 standard deviations below the mean in a standard normal distribution.
Step 2: Recall that the standard normal distribution is symmetric and centered at zero. Negative z-values correspond to the left side of the distribution.
Step 3: Identify the area under the curve to the left of z = -2.37. This area represents the cumulative probability (P-value) for z = -2.37.
Step 4: Match the graph provided with the z-statistic. The shaded region on the graph corresponds to the cumulative probability for z = -2.37, which is the area to the left of this z-value.
Step 5: To calculate the P-value, you would use a z-table or statistical software to find the cumulative probability associated with z = -2.37. This value represents the proportion of the distribution that lies to the left of z = -2.37.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Z-Score

A z-score indicates how many standard deviations an element is from the mean of a distribution. In this context, a z-score of -2.37 suggests that the value is 2.37 standard deviations below the mean, which is crucial for determining the corresponding area under the normal curve.
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Z-Scores From Given Probability - TI-84 (CE) Calculator

P-Value

The p-value represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. It is used to determine the significance of results in hypothesis testing, with lower p-values indicating stronger evidence against the null hypothesis.
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Step 3: Get P-Value

Normal Distribution

The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, defined by its mean and standard deviation. Understanding this distribution is essential for interpreting z-scores and p-values, as it provides the framework for calculating probabilities and areas under the curve.
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Related Practice
Textbook Question

Stating Hypotheses In Exercises 11–16, the statement represents a claim. Write its complement and state which is H0 and which is Ha.


p < 0.45

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Textbook Question

In Exercises 7–12, find the critical value(s) and rejection region(s) for the type of chi-square test with sample size n and level of significance α.


Two-tailed test, n=81,α=0.10

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Textbook Question

In Exercises 13–18, test the claim about the population mean μ at the level of significance α. Assume the population is normally distributed.

Claim: μ≥8000; α=0.01. Sample statistics: x_bar=77,000, s=450, n=25

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Textbook Question

Hypothesis Testing Using a P-Value In Exercises 33–38,

         

a. identify the claim and state and .

b. find the standardized test statistic z.

c. find the corresponding P-value.

d. decide whether to reject or fail to reject the null hypothesis.

e. interpret the decision in the context of the original claim.


MCAT Scores A random sample of 100 medical school applicants at a university has a mean total score of 505 on the MCAT. According to a report, the mean total score for the school’s applicants is more than 503. Assume the population standard deviation is 10.6. At alpha=0.01, is there enough evidence to support the report’s claim?

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Textbook Question

Stating the Null and Alternative Hypotheses In Exercises 25–30, write the claim as a mathematical statement. State the null and alternative hypotheses, and identify which represents the claim.


Attendance An amusement park claims that the mean daily attendance at the park is at least 20,000 people.

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Textbook Question

How do the requirements for a chi-square test for a variance or standard deviation differ from a z-test or a t-test for a mean?

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