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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. Assume all assumptions are met. In a recent year, the most popular colors for light trucks were white, \(31 \%\); black, \(19 \%\); silver \(11 \%\); red \(11 \%\); gray \(10 \%\); blue \(8 \%\); and other \(10 \%\). A survey of randomly selected light truck owners in a particular area revealed the following. At \(\alpha=0.05,\) do the proportions differ from those stated? $$ \begin{array}{ccccccc} \text { White } & \text { Black } & \text { Silver } & \text { Red } & \text { Gray } & \text { Blue } & \text { Other } \\ \hline 45 & 32 & 30 & 30 & 22 & 15 & 6 \end{array} $$

Short Answer

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

Step by step solution

01

State the Hypotheses

To determine if the proportions differ from those stated, we start by setting up our hypotheses. - Null Hypothesis: The proportions of light truck colors do not differ from those specified. Specifically, \( p_\text{White} = 0.31, \) \( p_\text{Black} = 0.19, \) \( p_\text{Silver} = 0.11, \) \( p_\text{Red} = 0.11, \) \( p_\text{Gray} = 0.10, \) \( p_\text{Blue} = 0.08, \) \( p_\text{Other} = 0.10 \). - Alternative Hypothesis: At least one of these proportions is different from the specified values. The claim is the alternative hypothesis.
02

Find the Critical Value

This is a test for goodness-of-fit using the Chi-square distribution, with \(k-1\) degrees of freedom, where \(k\) is the number of categories. Here, \(k=7\) (White, Black, Silver, Red, Gray, Blue, Other) thus the degrees of freedom are \(7 - 1 = 6\). At a significance level of \(\alpha=0.05\), the critical value from Chi-square distribution tables is approximately 12.592.
03

Compute the Test Value

First, calculate the expected frequencies based on the stated proportions and total sample size \(N = 180\):\[\begin{align*}E_\text{White} &= 0.31 \times 180 = 55.8, \E_\text{Black} &= 0.19 \times 180 = 34.2, \E_\text{Silver} &= 0.11 \times 180 = 19.8, \E_\text{Red} &= 0.11 \times 180 = 19.8, \E_\text{Gray} &= 0.10 \times 180 = 18, \E_\text{Blue} &= 0.08 \times 180 = 14.4, \E_\text{Other} &= 0.10 \times 180 = 18. \\end{align*}\]Now, use the formula for the Chi-square test statistic: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Compute this for each category, sum the values, and you get\[\chi^2 = \frac{(45-55.8)^2}{55.8} + \frac{(32-34.2)^2}{34.2} + \frac{(30-19.8)^2}{19.8} + \cdots + \frac{(6-18)^2}{18} = 15.98.\]
04

Make the Decision

Compare the test statistic value \( \chi^2 = 15.98 \) with the critical value 12.592. Since 15.98 > 12.592, we reject the null hypothesis.
05

Summarize the Results

There is sufficient evidence at \( \alpha = 0.05 \) to conclude that the proportions of popular light truck colors in this area differ from the stated proportions. The observed distribution does not fit the claimed distribution.

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Key Concepts

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

Chi-square test
The chi-square test is a statistical method used to determine if there is a significant difference between expected and observed data. It is often employed when you have categorical data and want to assess how well it fits an expected distribution. For our exercise, it's essential to compare the observed frequencies of light truck colors against the expected frequencies to see if they align. The chi-square test statistic is calculated using the formula:
  • \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]
Here, \( O_i \) stands for the observed frequency, and \( E_i \) stands for the expected frequency.
The degrees of freedom, which is necessary for finding the critical value, is computed as the number of categories minus one. For this exercise, with seven color categories, the degrees of freedom are six. Using the significance level of \( \alpha = 0.05 \), we compare the computed chi-square statistic against the critical value from chi-square distribution tables. If our test statistic exceeds the critical value, we reject the null hypothesis.
Goodness-of-fit
The goodness-of-fit test is a form of chi-square test designed to see how well an observed distribution of data matches a theoretical or expected distribution. In other words, it tests how well the sample data fits a distribution from a broader population.
In our exercise, we use the goodness-of-fit test to check if the observed proportions of truck colors correspond to the expected proportions provided in the problem statement.
  • White: 31%
  • Black: 19%
  • Silver: 11%
  • Red: 11%
  • Gray: 10%
  • Blue: 8%
  • Other: 10%
By comparing observed data from our sample of 180 trucks to these expected proportions, we determine if the real-world data fits the hypothetical distribution.
The chi-square statistic is a crucial element in determining the level of goodness-of-fit, as it highlights discrepancies between observed and expected values. If the calculated chi-square statistic is larger than the critical value, the fit is not good, leading us to reject the null hypothesis.
Null Hypothesis
The null hypothesis is a fundamental concept in statistical hypothesis testing. It acts as a starting point or assumption that there is no effect or no difference in the data, which we seek to test.
In the context of the exercise, the null hypothesis states that the distribution of light truck colors does not differ from the expected theoretical proportions:
  • \( p_\text{White} = 0.31 \)
  • \( p_\text{Black} = 0.19 \)
  • \( p_\text{Silver} = 0.11 \)
  • \( p_\text{Red} = 0.11 \)
  • \( p_\text{Gray} = 0.10 \)
  • \( p_\text{Blue} = 0.08 \)
  • \( p_\text{Other} = 0.10 \)
This implies that there’s no significant discrepancy between the observed sample data and the expected distribution. The aim of conducting our test is to determine whether to continue believing this hypothesis or to reject it based on evidence from our sample.
Alternative Hypothesis
The alternative hypothesis is in direct opposition to the null hypothesis and represents what we are trying to establish or prove. When conducting a chi-square test for goodness-of-fit, the alternative hypothesis suggests that the observed distribution of data does not fit the expected distribution.
For this exercise, the alternative hypothesis posits that at least one of the color proportions for light trucks in the stated area is different from the expected proportions:
  • For example, the actual proportion of white trucks may not be exactly 31%, as hypothesized.
Accepting the alternative hypothesis indicates a significant deviation between the observed and expected data.
If our test statistic is greater than the critical value, and we reject the null hypothesis, the claim is supported, confirming that differences in truck color proportions exist. This highlights the importance of the alternative hypothesis, as it directs us towards uncovering new insights about the underlying population.

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Most popular questions from this chapter

Select a three-digit state lottery number over a period of 50 days. Count the number of times each digit, 0 through 9 , occurs. Test the claim, at \(\alpha=0.05,\) that the digits occur at random.

When the expected frequency is less than 5 for a specific class, what should be done so that you can use the goodness-of-fit test?

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. Assume all assumptions are met. According to the Bureau of Transportation Statistics, on-time performance by the airlines is described as follows: $$ \begin{array}{lr} \text { Action } & \text { \% of Time } \\ \hline \text { On time } & 70.8 \\ \text { National Aviation System delay } & 8.2 \\ \text { Aircraft arriving late } & 9.0 \\ \text { Other (because of weather } & 12.0 \\ \quad \text { and other conditions) } & \end{array} $$ Records of 200 randomly selected flights for a major airline company showed that 125 planes were on time; 40 were delayed because of weather, 10 because of a National Aviation System delay, and the rest because of arriving late. At \(\alpha=0.05,\) do these results differ from the government's statistics?

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. Assume all assumptions are met. A survey was targeted at determining if educational attainment affected Internet use. Randomly selected shoppers at a busy mall were asked if they used the Internet and their highest level of education attained. The results are listed below. Is there sufficient evidence at the 0.05 level of significance that the proportion of Internet users differs for any of the groups? $$ \begin{array}{ccc} \begin{array}{c} \text { Graduated } \\ \text { college }+ \end{array} & \text { Attended college } & \text { Did not attend } \\ \hline 44 & 41 & 40 \end{array} $$

Perform the following steps. 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. Assume all assumptions are valid. A recent study of 100 individuals found the following living arrangement for men and women. The results are shown. Check the data for a dependent relationship at \(\alpha=0.05\). $$ \begin{array}{l|cccc} & \text { Spouse } & \text { Relative } & \text { Nonrelative } & \text { Alone } \\ \hline \text { Men } & 57 & 8 & 25 & 10 \\ \text { Women } & 53 & 5 & 28 & 14 \end{array} $$

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