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Chapter 15: Problem 9

The following hypotheses are given: \(H_{0}:\) Forty percent of the observations are in category \(A, 40\) percent are in \(B,\) and 20 percent are in \(\mathrm{C}\) \(H_{1}:\) The observations are not as described in \(H_{0}\). We took a sample of \(60,\) with the following results. $$ \begin{array}{|cc|} \hline \text { Category } & f_{o} \\ \hline \mathrm{A} & 30 \\ \mathrm{~B} & 20 \\ \mathrm{C} & 10 \\ \hline \end{array} $$ a. State the decision rule using the .01 significance level. b. Compute the value of chi-square. c. What is your decision regarding \(H_{0} ?\)

Short Answer

Expert verified
Decision: Fail to reject \(H_0\); insufficient evidence at 0.01 significance level.

Step by step solution

01

State the Decision Rule

To state the decision rule, we first need the degrees of freedom for the chi-square test. Degrees of freedom (df) is calculated as the number of categories minus 1. Here we have 3 categories (A, B, C), so \(df = 3 - 1 = 2\). Using a significance level of 0.01, look up the critical value of chi-square from the chi-square distribution table for 2 degrees of freedom, which is approximately 9.21. If the computed chi-square value exceeds this critical value, we reject \(H_0\).
02

Compute Expected Frequencies

Calculate the expected frequencies for each category based on \(H_0\). Since the total sample size is 60, the expected frequencies \(f_e\) are:- Category A: \(f_e = 0.4 \times 60 = 24\)- Category B: \(f_e = 0.4 \times 60 = 24\)- Category C: \(f_e = 0.2 \times 60 = 12\)
03

Calculate Chi-square Value

The chi-square statistic is calculated using the formula: \[\chi^2 = \sum \frac{(f_o - f_e)^2}{f_e}\]Substitute the observed \(f_o\) and expected \(f_e\) frequencies:- For A: \(\frac{(30 - 24)^2}{24} = \frac{36}{24} = 1.5\)- For B: \(\frac{(20 - 24)^2}{24} = \frac{16}{24} = 0.67\)- For C: \(\frac{(10 - 12)^2}{12} = \frac{4}{12} = 0.33\)Sum these values to get the chi-square statistic:\[\chi^2 = 1.5 + 0.67 + 0.33 = 2.5\]
04

Make a Decision Regarding H0

Compare the calculated chi-square value (2.5) to the critical value (9.21). Since 2.5 is less than 9.21, we fail to reject \(H_0\). This means there isn't sufficient evidence at the 0.01 significance level to conclude that the observations differ from the hypothesized distribution.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is crucial in deciding if data supports a specific claim or observation. In simple terms, it helps researchers assess whether their assumptions about a population hold true. In our exercise, two hypotheses are proposed: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)).
  • The null hypothesis (\(H_0\)) suggests that the distribution among categories A, B, and C is consistent with the stated proportions: 40%, 40%, and 20% respectively.
  • The alternative hypothesis (\(H_1\)) argues that the distribution does not follow this pattern.
Engaging in hypothesis testing involves analyzing sample data to confirm or disprove these proposed situations. In this setup, analysts often rely on statistical tests, like the Chi-Square Test, to examine differences between observed data and expected outcomes under \(H_0\). This procedure essentially scrutinizes whether the deviations between observed frequencies and expected frequencies are statistically significant or happen by chance.
Significance Level
The significance level is a predetermined threshold that helps decide whether to accept or reject the null hypothesis. It's often denoted by \(\alpha\) and is a critical component in hypothesis testing. In our problem, the significance level is set at 0.01, meaning a 1% chance of wrongly rejecting \(H_0\) when it is indeed true (Type I error).
  • A lower significance level, like 0.01, implies stricter criteria for rejecting \(H_0\), minimizing the likelihood of making an error that can lead to incorrect conclusions.
  • When calculating the chi-square statistic, it gets compared to a critical value derived from the chi-square distribution table. This table considers factors like degrees of freedom and the significance level (\(\alpha\)).
If the chi-square statistic exceeds the critical value, \(H_0\) is rejected, indicating the observations significantly deviate from the expected outcomes. In this scenario, with a chi-square result of 2.5 and a critical value of 9.21, the data lacks the necessary statistical evidence to reject \(H_0\) at a 0.01 level.
Degrees of Freedom
Degrees of freedom (df) is a term that refers to the number of independent values or quantities that can be assigned to a statistical distribution. For a chi-square test, it plays a central role in determining the critical value.
  • The degrees of freedom are calculated by subtracting 1 from the number of categories in the data. In our example:
    \[ df = ext{number of categories} - 1 = 3 - 1 = 2 \]
  • The greater the degrees of freedom, the more the distribution resembles a standard normal distribution over time.
Why does this matter? Because calculating df helps shape the chi-square distribution's critical value, which in turn affects whether \(H_0\) gets rejected. In our situation, a computation of 2 degrees of freedom aligns with a chi-square critical value of 9.21 for a 0.01 significance level. This allows the analysis of observed data to be conducted accurately, with careful consideration of any potential deviations from the expected frequencies.

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Most popular questions from this chapter

For many years TV executives used the guideline that 30 percent of the audience were watching each of the prime-time networks, that is \(\mathrm{ABC}, \mathrm{NBC}\) and \(\mathrm{CBS},\) and 10 percent were watching cable stations on a weekday night. A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder were viewing a cable station. At the .05 significance level, can we conclude that the guideline is still reasonable?

In the Tidewater Virginia TV market there are three commercial television stations, each with its own evening news program from 6: 00 to 6: 30 P.M. According to a report in this morning's local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11 ), and 33 on WSPD (channel 13). At the .05 significance level, is there a difference in the proportion of viewers watching the three channels?

A recent study by a large retailer designed to determine whether there was a relationship between the importance a store manager placed on advertising and the size of the store revealed the following sample information: $$ \begin{array}{|lrc|} \hline & \text { Important } & \text { Not Important } \\ \hline \text { Small } & 40 & 52 \\ \text { Medium } & 106 & 47 \\ \text { Large } & 67 & 32 \\ \hline \end{array} $$ What is your conclusion? Use the .05 significance level.

Two hundred managers employed in the lumber industry were randomly selected and interviewed regarding their concern about environmental issues. The response of each person was tallied into one of three categories: no concern, some concern, and great concern. The results were: $$ \begin{array}{|lccc|} \hline \begin{array}{l} \text { Level of } \\ \text { Management } \end{array} & \begin{array}{c} \text { No } \\ \text { Concern } \end{array} & \begin{array}{c} \text { Some } \\ \text { Concern } \end{array} & \begin{array}{c} \text { Great } \\ \text { Concern } \end{array} \\ \hline \text { Top management } & 15 & 13 & 12 \\ \text { Middle management } & 20 & 19 & 21 \\ \text { Supervisor } & 7 & 7 & 6 \\ \text { Group leader } & 28 & 21 & 31 \\ \hline \end{array} $$ Use the .01 significance level to determine whether there is a relationship between management level and environmental concern.

In a particular chi-square goodness-of-fit test there are six categories and 500 observations. Use the .01 significance level. a. How many degrees of freedom are there? b. What is the critical value of chi-square?
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