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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 7.5.4

Describe the difference between calculating the standardized test statistic, Z^2, for a chi-square test for variance and a chi-square test for standard deviation.

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Understand the purpose of the chi-square test: Both tests are used to assess whether the variance or standard deviation of a population matches a hypothesized value. The difference lies in the parameter being tested (variance vs. standard deviation).
Recall the formula for the chi-square test statistic: The test statistic is calculated as \( \chi^2 = \frac{(n-1) \cdot s^2}{\sigma^2} \), where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2 \) is the hypothesized population variance.
For a chi-square test for variance: Directly use the sample variance \( s^2 \) and hypothesized population variance \( \sigma^2 \) in the formula. The test statistic \( \chi^2 \) is calculated based on these values.
For a chi-square test for standard deviation: Convert the hypothesized standard deviation \( \sigma \) into variance by squaring it (\( \sigma^2 = \sigma \cdot \sigma \)). Similarly, ensure the sample standard deviation \( s \) is squared to obtain \( s^2 \). Then proceed with the same formula for \( \chi^2 \).
Interpretation: While the calculation process is similar, the key difference is whether the hypothesized and sample values are expressed in terms of variance or standard deviation. Always ensure consistency in units (variance or standard deviation) before applying the formula.

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Key Concepts

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

Chi-Square Test for Variance

The chi-square test for variance assesses whether the variance of a population is equal to a specified value. It uses the chi-square statistic, which is calculated by comparing the observed variance from a sample to the expected variance under the null hypothesis. This test is particularly useful in quality control and research to determine if variability in data is consistent with a theoretical model.
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Chi-Square Test for Standard Deviation

The chi-square test for standard deviation is similar to the test for variance but focuses specifically on the standard deviation. Since the standard deviation is the square root of the variance, the test evaluates whether the sample standard deviation significantly differs from a hypothesized population standard deviation. This distinction is important in contexts where standard deviation is a more intuitive measure of spread than variance.
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Calculating Standard Deviation

Standardized Test Statistic (Z^2)

The standardized test statistic, often denoted as Z^2 in the context of chi-square tests, is a measure that allows for the comparison of the observed data to the expected data under the null hypothesis. It is calculated by taking the difference between observed and expected values, squaring it, and normalizing by the expected values. This statistic helps determine the significance of the results, indicating whether to reject the null hypothesis based on the calculated p-value.
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Step 2: Calculate Test Statistic
Related Practice
Textbook Question

Graphical Analysis In Exercises 21 and 22, state whether each standardized test statistic z allows you to reject the null hypothesis. Explain your reasoning.


a. z = -1.301

b. z = 1.203

c. z = 1.280

d. z = 1.286


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Textbook Question

Deciding on a Distribution In Exercises 31 and 32, decide whether you should use the standard normal sampling distribution or a t-sampling distribution to perform the hypothesis test. Justify your decision. Then use the distribution to test the claim. Write a short paragraph about the results of the test and what you can conclude about the claim.


Tuition and Fees An education publication claims that the mean in-state tuition and fees at public four-year institutions by state is more than \$10,500 per year. A random sample of 30 states has a mean in-state tuition and fees at public four-year institutions of \$10,931 per year. Assume the population standard deviation is \$2380. At α=0.01, test the publication’s claim.

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Textbook Question

Identifying a Test In Exercises 21–24, determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.


Ha: σ^2 = 142

H0: σ ≠ 142

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Textbook Question

Identifying Type I and Type II Errors In Exercises 31–36, describe type I and type II errors for a hypothesis test of the indicated claim.


Phone Repairs A cell phone repair shop advertises that the mean cost of repairing a phone screen is less than \$120.

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Textbook Question

Writing You are testing a claim and incorrectly use the standard normal sampling distribution instead of the t-sampling distribution, mistaking the sample standard deviation for the population standard deviation. Does this make it more or less likely to reject the null hypothesis? Is this result the same no matter whether the test is left-tailed, right-tailed, or two-tailed? Explain your reasoning.

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Textbook Question

The mean of a random sample of 18 test scores is x_bar. The standard deviation of the population of all test scores is sigma= 6. Under what condition can you use a z-test to decide whether to reject a claim that the population mean is mu=88?

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