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For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Reading Program Summer reading programs are very popular with children. At the Citizens Library, Team Ramona read an average of 23.2 books with a standard deviation of \(6.1 .\) There were 21 members on this team. Team Beezus read an average of 26.1 books with a standard deviation of \(2.3 .\) There 23 members on this team. Did the variances of the two teams differ? Use \(\alpha=0.05 .\)

Short Answer

Expert verified
The variances differ; reject the null hypothesis at \\(\alpha=0.05\\).

Step by step solution

01

State the Hypotheses and Identify the Claim

We want to determine if there is a difference in the variances of the two teams' book readings. Thus, we'll set up our hypotheses as follows:* Null Hypothesis \(H_0\): The variances of the number of books read by the two teams are equal, \(\sigma_1^2 = \sigma_2^2\).* Alternative Hypothesis \(H_a\): The variances of the number of books read by the two teams are not equal, \(\sigma_1^2 eq \sigma_2^2\).The claim is the alternative hypothesis.
02

Find the Critical Value

Since we are comparing variances, we'll use the F-distribution. The degrees of freedom (df) for Team Ramona is \(df_1 = n_1 - 1 = 21 - 1 = 20\), and for Team Beezus is \(df_2 = n_2 - 1 = 23 - 1 = 22\).The critical value for an F-test at \(\alpha = 0.05\) for a two-tailed test with \(df_1 = 20\) and \(df_2 = 22\) can be found using an F-table or calculator. We'll consider both tails, i.e., find \(F_{\alpha/2}\) and \(F_{1-\alpha/2}\).For this calculation, the critical values are approximately \(F_{0.025} = 2.54\) and \(F_{0.975} = 1/2.54 = 0.39\).
03

Compute the Test Value

The test statistic for comparing the variances is given by the formula:\[ F = \frac{s_1^2}{s_2^2} \]Where \(s_1 = 6.1\) is the standard deviation for Team Ramona and \(s_2 = 2.3\) is the standard deviation for Team Beezus.First, find the variances:* Team Ramona: \(s_1^2 = 6.1^2 = 37.21\)* Team Beezus: \(s_2^2 = 2.3^2 = 5.29\)Compute the F-statistic:\[ F = \frac{37.21}{5.29} \approx 7.03 \]
04

Make the Decision

Compare the calculated F value to the critical values from the F-distribution. We have:- Calculated \(F = 7.03\).- Critical value range: \(0.39 < F < 2.54\).Since \(F = 7.03\) is greater than the upper critical value of \(2.54\), we reject the null hypothesis \(H_0\).
05

Summarize the Results

Since the calculated F value lies outside the range of the critical values, we reject the null hypothesis, indicating that there is sufficient evidence to conclude that the variances in the number of books read by Team Ramona and Team Beezus are significantly different at a 0.05 significance level.

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Key Concepts

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

F-distribution
The F-distribution is a fundamental concept in statistics, especially when comparing variances of two different data sets. It is used extensively in experiments that aim to test hypotheses about whether two population variances are equal. The distribution is named after Sir Ronald Fisher, and it is a probability distribution that arises in the testing of whether two estimated variances are equal.

When you perform an F-test, you derive an F-statistic calculated by the ratio \left( F = \frac{s_1^2}{s_2^2} \right), where \(s_1^2\) and \(s_2^2\) are the sample variances. The F-distribution is asymmetrical and stretches out more on the side of increasing variance ratios.

This distribution depends on degrees of freedom related to the two data samples being compared. Essentially, the shape of the F-distribution will vary based on these degrees of freedom, specifically \(df_1\) for the first group and \(df_2\) for the second group. In our example, Team Ramona has \(df_1 = 20\) and Team Beezus has \(df_2 = 22\). The F-distribution allows us to set a range of critical values, within which the null hypothesis is accepted.

To understand if the variances differ, you'll compare your calculated F-statistic with this range of critical values.
Critical value
Critical values are essential in hypothesis testing as they determine the threshold at which you decide whether to reject the null hypothesis. When working with the F-distribution, the critical value enables you to evaluate if your test statistic falls within the acceptance region or not.

Critical values for the F-distribution can be found using F-distribution tables or statistical software. It is important to choose the correct tail based on the nature of your test. In a two-tailed test, you will find two critical values — one at the lower end and one at the upper end. These define the acceptance region for the null hypothesis.
  • If the calculated F-statistic falls below the lower critical value or above the upper critical value, the null hypothesis is rejected.
  • If it lies within, the null hypothesis cannot be rejected.

For this exercise, using \(\alpha = 0.05\), the critical values were \(F_{0.025} = 2.54\) and \(F_{0.975} = 0.39\). Since the calculated F-statistic (7.03) is greater than 2.54, it falls outside the acceptance region, leading to the rejection of the null hypothesis.
Null Hypothesis
In hypothesis testing, the null hypothesis represents the default assumption that there is no effect or difference. It is denoted as \(H_0\). When comparing variances, \(H_0\) typically asserts that the variances are equal.

For our exercise, the null hypothesis \(H_0\) states: The variances of the number of books read by Team Ramona and Team Beezus are equal, or mathematically, \(\sigma_1^2 = \sigma_2^2\).

The purpose of testing the null hypothesis is to challenge this assumption and determine if there is significant evidence to suggest otherwise. If the computed test statistic falls into a critical region determined by the F-distribution, it provides evidence against \(H_0\).

In our example, rejecting \(H_0\) indicates significant evidence that the variances for Team Ramona and Team Beezus differ.
Alternative Hypothesis
The alternative hypothesis is the statement you aim to provide evidence for in a hypothesis test. It represents a new effect or a difference, contrasted with the null hypothesis. It is typically denoted as \(H_a\), and in many cases, aligns with the research question or claim.

In this context, the alternative hypothesis \(H_a\) posits that the variances between the two teams are not equal; mathematically, \(\sigma_1^2 eq \sigma_2^2\).
  • The alternative hypothesis is your claim which you suspect to be true based on your observations.
  • Evidence in favor of \(H_a\) leads to the rejection of \(H_0\).

In our example, the calculated F-statistic provides sufficient evidence to support the alternative hypothesis, concluding that the variances in the number of books read between Team Ramona and Team Beezus differ significantly. This supported the claim and stood against the null hypothesis under the significance level of \(\alpha = 0.05\).

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Most popular questions from this chapter

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax- exempt properties for the two random samples are shown. The data are given in millions of dollars. At \(\alpha=0.05,\) is there enough evidence to support the tax collector's claim that the means are different? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Winter Temperatures A random sample of daily high temperatures in January and February is listed. At \(\alpha=0.05,\) can it be concluded that there is a difference in variances in high temperature between the two months? $$ \begin{array}{l|ccccccccccc}{\text { Jan. }} & {31} & {31} & {38} & {24} & {24} & {42} & {22} & {43} & {35} & {42} \\ \hline \text { Feb. } & {31} & {29} & {24} & {30} & {28} & {24} & {27} & {34} & {27}\end{array} $$

Heights of 9 -Year-Olds At age 9 the average weight \((21.3 \mathrm{kg})\) and the average height \((124.5 \mathrm{cm})\) for both boys and girls are exactly the same. A random sample of 9 -year-olds yielded these results. At \(\alpha=0.05,\) do the data support the given claim that there is a difference in heights? $$ \begin{array}{lcc}{} & {\text { Boys }} & {\text { Girls }} \\ \hline \text { Sample size } & {60} & {50} \\ {\text { Mean height, } \mathrm{cm}} & {123.5} & {126.2} \\ {\text { Population variance }} & {98} & {120}\end{array} $$

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Cholesterol Levels A medical researcher wishes to see if he can lower the cholesterol levels through diet in 6 people by showing a film about the effects of high cholesterol levels. The data are shown. At \(\alpha=0.05,\) did the cholesterol level decrease on average? $$ \begin{array}{lllllll}\hline \text { Patient } & {1} & {2} & {3} & {4} & {5} & {6} \\ \hline \text { Before } & {243} & {216} & {214} & {222} & {206} & {219} \\ \hline \text { After } & {215} & {202} & {198} & {195} & {204} & {213}\end{array} $$

What are the characteristics of the \(F\) distribution?

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