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The table shows the results of rolling a six-sided die 120 times. $$\begin{array}{|c|c|}\hline \text { Outcome on Die } & \text { Frequency } \\\\\hline 1 & 27 \\\\\hline 2 & 20 \\\\\hline 3 & 22 \\ \hline 4 & 23 \\\\\hline 5 & 19 \\\\\hline 6 & 9 \\\\\hline\end{array}$$ Test the hypothesis that the die is not fair. A fair die should produce equal numbers of each outcome. Use the four-step procedure with a significance level of \(0.05\), and state your conclusion clearly.

Short Answer

Expert verified
Following the four-step procedure for hypothesis testing, it is concluded that there is not enough evidence to reject the null hypothesis. Thus, the die can be considered as fair at a significance level of 0.05.

Step by step solution

01

State the Hypothesis

Null hypothesis (\(H_0\) ): The die is fair, i.e., the distribution is uniform. Alternate hypothesis (\(H_a\)): The die is not fair, i.e., the distribution is not uniform.
02

Formulate an Analysis plan

To evaluate the hypotheses, it will be used a Chi-Square test for goodness of fit which tests if the outcomes of the die fits a uniform distribution. This test will compare the observed frequencies to the expected frequencies. The test statistic \(\chi^2\) is calculated by the formula, \(\chi^2 = \sum \[(O_i - E_i)^2/E_i]\), where \(O_i\) and \(E_i\) are observed and expected frequencies, respectively.
03

Conduct the Analysis

Calculate the expected outcomes and compute value of \(\chi^2\) :Expected frequency for each outcome = Total trials/Number of outcomes = 120 / 6 = 20\(\chi^2 = (27-20)^2/20 + (20-20)^2/20 + (22-20)^2/20 + (23-20)^2/20 + (19-20)^2/20 + (9-20)^2/20 = 2.45 \)
04

Evaluate the Hypothesis

With 5 degrees of freedom (6 outcomes - 1), and at a significance level of 0.05, the critical value from the Chi-Square distribution table is approximately 11.07. Since the calculated test statistic 2.45 is less than the critical value, it is not in the rejection region. Thus, there is not enough evidence to reject the null hypothesis.

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Key Concepts

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

Statistical Hypothesis Testing
Statistical hypothesis testing is a method of decision-making in the context of statistical analysis. It involves making assumptions, or hypotheses, about a population parameter and then determining whether the data collected supports these assumptions. To conduct a statistical test, specific steps must be followed:

  • State the Hypotheses: Firstly, you articulate the null hypothesis ((H_0)) and alternate hypothesis ((H_a)). The null hypothesis typically represents the 'status quo' or a condition of no effect or no difference. In the context of dice, we assume that the die is fair, i.e., all outcomes are equally likely.
  • Choose the Appropriate Test: Depending on the nature of your data and hypotheses, you choose a test. For categorical data like dice rolls, you often use the Chi-Square test for goodness of fit.
  • Calculate the Test Statistic: This involves using a formula that incorporates the observed and expected frequencies of the data to compute a test statistic value.
  • Make a Decision: You compare the test statistic to a critical value from the relevant distribution table, considering the significance level, to decide if the null hypothesis can be rejected or not.
The essence of hypothesis testing lies in determining the probability that the observed data could occur by chance if the null hypothesis were true. If this probability is very low (usually less than the significance level), the null hypothesis is rejected.
Null Hypothesis
The null hypothesis, denoted as (H_0), is a fundamental concept in statistics that represents a default position: that there is no effect or no difference. In the exercise, the null hypothesis is that the die is fair, and each face should have the same probability of showing up after a roll. The alternate hypothesis (H_a) contradicts this and suggests that the die is not fair.

In statistical testing, the objective is often to provide evidence against the null hypothesis. To do this, researchers collect data and calculate a test statistic. If the test statistic is extreme and falls into the tail of the distribution (usually beyond the critical value), the null hypothesis is considered unlikely. However, it is crucial to understand that failing to reject the null hypothesis does not necessarily mean it is true - it suggests that based on the data, there is insufficient evidence to support the alternate hypothesis.
Degrees of Freedom
Degrees of freedom (df) is a concept that describes the number of independent values in a calculation that can vary without violating any imposed restrictions. In the context of the Chi-Square test for goodness of fit, the degrees of freedom are calculated as the number of categories minus one (number of outcomes - 1).

In the dice exercise, there are six potential outcomes (the faces of the die), so the degrees of freedom are five (6 - 1). The concept of degrees of freedom is crucial when it comes to determining the critical value from the Chi-Square distribution table, as it affects which value to use for making the decision on whether to reject the null hypothesis. If the test statistic is below the critical value for the corresponding degrees of freedom and significance level, the null hypothesis is not rejected. Bigger degrees of freedom correspond to wider and flatter Chi-Square distributions, which can affect the decision-making threshold in hypothesis testing.

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Most popular questions from this chapter

In the study referenced in exercise \(10.33\), researchers also collected data on use of apps to monitor diet and calorie intake. The data are reported in the table. Test the hypothesis that diet app use and gender are associated. Use a \(0.05\) significance level. $$ \begin{array}{ccc} \text { Use } & \text { Male } & \text { Female } \\ \text { Yes } & 43 & 241 \\ \text { No } & 50 & 84 \\ \hline\end{array}$$

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In a 2015 study reported in the Journal of American College Health, Cho. et al. surveyed college students on their use of apps to monitor their exercise and fitness. The data are reported in the table. Test the hypothesis that fitness app use and gender are associated. Use a \(0.05\) significance level. See page 552 for guidance. $$\begin{array}{|ccc|}\hline \text { Use } & \text { Male } & \text { Female } \\\ \hline \text { Yes } & 84 & 268 \\ \hline \text { No } & 9 & 57 \\ \hline\end{array}$$

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