/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A freeway with four lanes in eac... [FREE SOLUTION] | 91Ó°ÊÓ

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A freeway with four lanes in each direction was studied to see whether drivers prefer to drive on the inside lanes. A total of 1000 automobiles were observed during heavy earlymorning traffic, and the number of cars in each lane was recorded: $$ \begin{array}{l|llll} \text { Lane } & 1 & 2 & 3 & 4 \\ \hline \text { Observed Count } & 294 & 276 & 238 & 192 \end{array} $$ Do the data present sufficient evidence to indicate that some lanes are preferred over others? Test using \(\alpha=.05 .\) If there are any differences, discuss the nature of the differences.

Short Answer

Expert verified
If so, which lanes are more popular? Answer: Yes, there is evidence to suggest that drivers have a preference for driving on specific lanes of a freeway. Specifically, lanes 1 and 2 appear to be more popular than lanes 3 and 4, as they have higher observed counts than the expected count in a uniform distribution.

Step by step solution

01

Formulate the null and alternative hypotheses

The null hypothesis (H0) states that there is no preference for driving on any specific lane, which implies that each lane has an equal chance of being selected by the drivers. The alternative hypothesis (H1) states that some lanes are preferred over others. - Null Hypothesis (\(H_0\)): The lane preference distribution is uniform. - Alternative Hypothesis (\(H_1\)): The lane preference distribution is not uniform.
02

Set up the significance level and degrees of freedom

The significance level is given in the problem statement: \(\alpha = 0.05\). The degrees of freedom (df) can be calculated by the formula: df = (number of lanes - 1) = 4 - 1 = 3.
03

Calculate the expected frequency for each lane

Under the null hypothesis, the expected frequency of cars in each lane would be equal. With 1000 observed cars, the expected frequency for all lanes would be: Expected Count: \(E_i = \frac{\text{total observed counts}}{n} = \frac{1000}{4} = 250\)
04

Calculate the chi-squared test statistic

The chi-squared test statistic is calculated as follows: $$X^2 = \sum\limits_{i=1}^4 \cfrac{(O_i - E_i)^2}{E_i}$$ where i represents the lane number, \(O_i\) is the observed count for each lane, and \(E_i\) is the expected count under the null hypothesis. Using the observed and expected frequencies, we get: $$X^2 = \cfrac{(294-250)^2}{250} + \cfrac{(276-250)^2}{250} + \cfrac{(238-250)^2}{250} + \cfrac{(192-250)^2}{250} \approx 16.02$$
05

Determine the p-value

Using the chi-squared distribution table or a calculator, we find the p-value for the test statistic. Since we have df = 3, we will find the p-value corresponding to \(X^2 = 16.02\). The p-value is 0.00063 which is less than our chosen significance level (\(\alpha = 0.05\)).
06

Conclusion

Since the p-value is less than the significance level, we reject the null hypothesis, concluding that there is enough evidence to suggest that there is an uneven distribution of cars among the lanes. In other words, some lanes are indeed preferred over others. Specifically, lanes 1 and 2 appear to be more popular than lanes 3 and 4 as they have higher observed counts than the expected count in a uniform distribution.

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Key Concepts

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

Hypothesis Testing
Understanding hypothesis testing is crucial when dealing with statistical data analysis. It is a systematic method used to evaluate assumptions about a dataset. In the context of our exercise, we determine if there is a preference for certain lanes on a freeway. We start by assuming that no preference exists, which leads us to establish a null hypothesis. We then collect data, calculate the necessary statistics, and reach a conclusion based on this evidence. The process includes setting a significance level which acts as the benchmark for deciding whether the observed data is sufficiently unlikely under the null hypothesis. If the test statistic calculated from the data falls beyond this benchmark, we reject the null hypothesis in favor of the alternative hypothesis. This clear-cut decision-making process is guided by the principles of hypothesis testing.
Null and Alternative Hypotheses
The formulation of null (\(H_0\)) and alternative (\(H_1\)) hypotheses is a foundational step in hypothesis testing. The null hypothesis represents the default assumption that there is no effect or no difference, proposing that lane preference is uniform across all lanes in our example. Alternatively, the alternative hypothesis posits that what we expect to prove is true—in this case, that there is a lane preference. These hypotheses are mutually exclusive. During analysis, we seek evidence against the null hypothesis. By doing so, the test's outcome either entails rejecting the null hypothesis, if the evidence is strong, or failing to reject it, if the evidence is weak. This dichotomy encapsulates the possible conclusions we can reach based on our data and statistical tests.
Chi-squared Distribution
The chi-squared distribution is a fundamental concept in statistics, especially in the context of goodness-of-fit tests like the one described in our freeway lane preference study. It is used to compare observed data with data we would expect to obtain according to a specific hypothesis. The chi-squared distribution emerges when we sum the squares of independent standard normal random variables. In hypothesis testing, the chi-squared statistic measures how our observed data diverges from the expected data under the null hypothesis. Larger values of the chi-squared statistic indicate greater divergence. The shape of the chi-squared distribution is characterized by degrees of freedom, which in our case is determined by the number of categories minus one. This distribution is crucial in determining the p-value, which helps us understand whether the observed results are due to chance or if they reflect a significant deviation from the null hypothesis.
Significance Level
In hypothesis testing, the significance level (\(\alpha\)) is a threshold that determines when to reject the null hypothesis. Conventionally set at 0.05, or 5%, this level indicates the probability of making a Type I error—rejecting a true null hypothesis. When we set a significance level, we are essentially defining what we would consider an unlikely result if the null hypothesis were true. In our example, a significance level of 0.05 signifies that we would expect our test statistic to fall into this extreme 5% area by chance alone only 5 times out of 100. If our calculated p-value is less than the significance level, the result is statistically significant, leading us to reject the null hypothesis and accept the alternative. The smaller the p-value, the stronger the evidence against the null hypothesis. Choosing an appropriate significance level is crucial as it reflects the degree of certainty one wishes to have before drawing a conclusion about the hypothesis.

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Most popular questions from this chapter

In 2006 , a new law passed in Massachusetts would require all residents to have health insurance. Low-income residents would get state subsidies to help pay insurance premiums, but everyone would pay something for health services. The plan would penalize people without any insurance and charge fees to employers who don't provide coverage. An ABC News/Washington Post poll \(^{4}\) involving \(n=1027\) adults nationwide asked the question, "Would you support or oppose this plan in your state?" The data that follows is based on the results of this study. a. Are there significant differences in the proportions of those surveyed who support, oppose, and are unsure about this plan among Democrats, Independents, and Republicans? Use \(\alpha=.05 .\) b. If significant differences exist, describe the nature of the differences by finding the proportions of those who support, oppose, and are unsure for each of the given affiliations.

Suppose that a consumer survey summarizes the responses of \(n=307\) people in a contingency table that contains three rows and five columns. How many degrees of freedom are associated with the chi-square test statistic?

Is your holiday turkey safe? A "new federal survey found that \(13 \%\) of turkeys are contaminated with the salmonella bacteria responsible for 1.3 million illnesses and about 500 deaths in a year in the US."15 Use the table that follows to determine if there is a significant difference in the contamination rate at three processing plants. One hundred turkeys were randomly selected from each of the processing lines at these three plants. Is there a significant difference in the rate of salmonella contamination among these three processing plants? If there is a significant difference, describe the nature of these differences. Use \(\alpha=.01\).

Use Table 5 in Appendix I to find the value of \(\chi^{2}\) with the following area \(\alpha\) to its right: a. \(\alpha=.05, d f=3\) b. \(\alpha=.01, d f=8\)

Suppose you are interested in following two independent traits in snap peas- seed texture \((\mathrm{S}=\) smooth, \(\mathrm{s}=\) wrinkled \()\) and seed color \((\mathrm{Y}=\) yellow, \(\mathrm{y}=\) green \()-\) in a second-generation cross of heterozygous parents. Mendelian theory states that the number of peas classified as smooth and yellow, wrinkled and yellow, smooth and green, and wrinkled and green should be in the ratio \(9: 3: 3: 1 .\) Suppose that 100 randomly selected snap peas have \(56,\) \(19,17,\) and 8 in these respective categories. Do these data indicate that the 9: 3: 3: 1 model is correct? Test using \(\alpha=.01\)

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