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State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test the following claims: a. The variances of populations \(A\) and \(B\) are not equal. b. The standard deviation of population I is larger than the standard deviation of population II. c. The ratio of the variances for populations \(A\) and \(B\) is different from 1. d. The variability within population \(\mathrm{C}\) is less than the variability within population D.

Short Answer

Expert verified
a. \(H_{0}\): \(\sigma_{A}^2 = \sigma_{B}^2\), \(H_{a}\): \(\sigma_{A}^2 \neq \sigma_{B}^2\). b. \(H_{0}\): \(\sigma_{I} \leq \sigma_{II}\), \(H_{a}\): \(\sigma_{I} > \sigma_{II}\). c. \(H_{0}\): \(\frac{\sigma_{A}^2}{\sigma_{B}^2} = 1\), \(H_{a}\): \(\frac{\sigma_{A}^2}{\sigma_{B}^2} \neq 1\). d. \(H_{0}\): \(\sigma_{C}^2 \geq \sigma_{D}^2\), \(H_{a}\): \(\sigma_{C}^2 < \sigma_{D}^2\).

Step by step solution

01

Formulate Hypotheses for Claim a

Claim a states that the variances of populations A and B are not equal. Therefore, \(H_{0}\): The variances of populations A and B are equal (\(\sigma_{A}^2 = \sigma_{B}^2\)). \(H_{a}\): The variances of populations A and B are not equal (\(\sigma_{A}^2 \neq \sigma_{B}^2\)).
02

Formulate Hypotheses for Claim b

Claim b states that the standard deviation of population I is larger than the standard deviation of population II. Therefore, \(H_{0}\): The standard deviation of population I is less than or equal to the standard deviation of population II (\(\sigma_{I} \leq \sigma_{II}\)). \(H_{a}\): The standard deviation of population I is larger than the standard deviation of population II (\(\sigma_{I} > \sigma_{II}\)).
03

Formulate Hypotheses for Claim c

Claim c states that the ratio of the variances for populations A and B is different from 1. Therefore, \(H_{0}\): The ratio of the variances for populations A and B is equal to 1 (\(\frac{\sigma_{A}^2}{\sigma_{B}^2} = 1\)). \(H_{a}\): The ratio of the variances for populations A and B is not equal to 1 (\(\frac{\sigma_{A}^2}{\sigma_{B}^2} \neq 1\)).
04

Formulate Hypotheses for Claim d

Claim d states that the variability within population C is less than the variability within population D. Therefore, \(H_{0}\): The variability within population C is greater than or equal to the variability within population D (\(\sigma_{C}^2 \geq \sigma_{D}^2\)). \(H_{a}\): The variability within population C is less than the variability within population D (\(\sigma_{C}^2 < \sigma_{D}^2\)).

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental part of statistical hypothesis testing. It represents a statement of no effect or no difference. When we perform a hypothesis test, the null hypothesis often suggests that any observed difference in the sample is due to random chance. It is often denoted as \(H_{0}\).

For example, in testing whether the variances of two populations are equal, the null hypothesis would be that these variances are indeed the same. Mathematically, we can represent this null hypothesis for variances of populations A and B as \(\sigma_{A}^2 = \sigma_{B}^2\).

The null hypothesis acts as a starting point for statistical testing, and it is tested against the alternative hypothesis. By initially assuming that the null hypothesis is true, we can calculate probabilities and determine whether our observed data is unlikely under this assumption.
Alternative Hypothesis
While the null hypothesis suggests no effect or no difference, the alternative hypothesis proposes what we really want to prove. It suggests the presence of an actual effect or difference. This hypothesis is usually denoted as \(H_{a}\).

In the case of testing variances, if we suspect that the variances of populations A and B are not equal, the alternative hypothesis would state \(\sigma_{A}^2 eq \sigma_{B}^2\).

The alternative hypothesis is crucial as it offers a direction for the test. When there is sufficient evidence to reject the null hypothesis, we may accept the alternative hypothesis instead. It provides a framework for the statistical evidence we gather.
Variance Comparison
Variance comparison is an essential part of many statistical tests, especially when comparing the distribution of data between two or more groups. Variance measures how far a set of numbers is spread out; it is a powerful way to understand variability within a population.

For example, when comparing populations A and B, we can test whether \(\sigma_{A}^2\) equals \(\sigma_{B}^2\). If the variances are significantly different, it might imply that the respective populations have differing levels of variability or spread in their data.

Such comparisons often lead to important insights, such as assessing the consistency or reliability of different methods, processes, or treatments. In statistical testing, comparing variances helps us understand the differences in data distributions at a deeper level.
Standard Deviation Comparison
Standard deviation is another measure of variability, often used in conjunction with variance, as it is the square root of variance. It provides an intuitive sense of how much individual data points deviate from the mean of a dataset.

When we compare the standard deviation of two populations, say I and II, we might test claims like \(\sigma_{I} > \sigma_{II}\). This comparison helps to determine which population has more consistency around the mean.

In hypothesis testing, leveraging standard deviation can give more context to variance comparisons. By comparing standard deviations, we gain insights into the relative homogeneity or heterogeneity of the datasets. Understanding standard deviations is key to making sense of variability and interpreting results within practical contexts.

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Most popular questions from this chapter

Determine the \(p\) -value that would be used to test the following hypotheses when \(z\) is used as the test statistic. a. \(\quad H_{o}: p_{1}=p_{2}\) versus \(H_{a}: p_{1}>p_{2},\) with \(z *=2.47\) b. \(\quad H_{o}: p_{A}=p_{B}\) versus, \(H_{a}: p_{A} \neq p_{B},\) with \(z \star=-1.33\) c. \(\quad H_{a}: p_{1}-p_{2}=0\) versus \(H_{a}: p_{1}-p_{2}<0,\) with \(z *=-0.85\) d. \(\quad H_{o}: p_{m}-p_{f}=0\) versus \(H_{a}: p_{m}-p_{f}>0,\) with \(z *=3.04\)

A Bloomberg News poll found that Americans plan to keep spending down over the next six months due to the uncertain economy (USA Today, September 17, 2009). Suppose a group of 15 households noted their household spending in March and then noted their household spending six months later in September. The mean monthly difference (former spending - current spending was calculated to be \(\$ 75.50\) with a standard deviation of \(\$ 66.20 .\) Does this sample of households show sufficient evidence of increased household savings? Use the 0.05 level of significance and assume normality of spending amounts.

Approximately \(95 \%\) of the sunflowers raised in the United States are grown in the states of North Dakota, South Dakota, and Minnesota. To compare yield rates between North and South Dakota, 11 sunflower-producing counties were randomly selected from North Dakota and 14 sunflower-producing counties were randomly selected from South Dakota. Their 2008 yields, in pounds per acre, are recorded below.$$\begin{array}{lllllll}\text { N. Dakota } & & & & & \\\\\hline 1296 & 1475 & 1573 & 1517 & 1242 & 1385 \\ 1128 & 1524 & 1644 & 1377 & 1270 & & \\\\\hline \text { S. Dakota } & & & & & \\\\\hline 1551 & 890 & 1710 & 1960 & 1988 & 1861 & 1870 \\ 1110 & 1674 & 1100 & 1381 & 2167 & 1130 & 1280 \\\\\hline\end{array}$$ Find the \(95 \%\) confidence interval for the difference between the mean sunflower yield for all North Dakota sunflower-producing counties and the mean sunflower yield for all South Dakota sunflower-producing counties. Assume normality of yield rates.

Many cheeses are produced in the shape of a wheel, and due to manufacturing inconsistencies, the amount of cheese, measured by weight, varies from wheel to wheel. Heidi Cembert wishes to determine if there is a significant difference, at the \(10 \%\) level, between the weight per wheel of Gouda and Brie cheese. She randomly samples 16 wheels of Gouda and finds the mean to be 1.2 pounds with a standard deviation of 0.32 pound and then samples 14 wheels of Brie and finds the mean to be 1.05 pounds with a standard deviation of 0.25 pound. At the 0.05 level of significance, is there sufficient evidence to support Heidi's contention that there is a difference in the mean weights of the two types of cheese?

Find the \(90 \%\) confidence interval for the difference between two means based on this information about two samples. Assume independent samples from normal populations.$$\begin{array}{lccc}\text { Sample } & \text { Number } & \text { Mean } & \text { Std. Dev. } \\\\\hline 1 & 20 & 35 & 22 \\\2 & 15 & 30 & 16 \\\\\hline\end{array}$$

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