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Explain why chi-square goodness-of-fit tests are always right tailed.

Short Answer

Expert verified
Chi-square goodness-of-fit tests are always right-tailed because the chi-square statistic is always non-negative, and we are interested in large deviations from the expected frequencies.

Step by step solution

01

Understand the Chi-Square Statistic

The chi-square goodness-of-fit test is used to determine if an observed frequency distribution fits an expected distribution. The test statistic is calculated as: \text - \text{statistic}\frac{(O_i - E_i)^2}{E_i} where \( O_i \) represents observed frequencies and \( E_i \) represents expected frequencies.
02

Positive Chi-Square Statistic

Notice that the formula for the chi-square statistic involves the square of \( (O_i - E_i) \). Since squaring any real number results in a non-negative value, the chi-square statistic is always non-negative.
03

Distribution Properties

The chi-square distribution is asymmetric and starts at zero, extending to positive infinity. It is defined only for non-negative values.
04

Right-Tailed Nature

In hypothesis testing, we are interested in whether the observed chi-square statistic is significantly higher than what we would expect under the null hypothesis. This involves the right tail of the distribution, where larger chi-square values suggest greater deviations from the expected distribution.
05

Final Conclusion

Given the properties of the chi-square distribution and the interest in detecting significant deviations, chi-square goodness-of-fit tests are always right-tailed.

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Key Concepts

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

Chi-Square Statistic
The chi-square statistic is a key component in the chi-square goodness-of-fit test. It helps determine if there is a significant difference between observed and expected frequencies. The formula for the chi-square statistic is \( \text{X}^2 = \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) represents the observed frequencies and \( E_i \) represents the expected frequencies. By squaring the difference between observed and expected frequencies, we ensure that all values are non-negative, as squaring any real number results in a positive value. The sum of these squared values divided by the expected frequencies results in the chi-square statistic.
Observed Frequency Distribution
The observed frequency distribution is a count of occurrences or instances of each category within your sample data. For example, if you are studying the color of cars in a parking lot and you count how many are red, blue, or green, those counts are your observed frequencies. This data will be compared against the expected distribution to see if there are any significant differences.
Expected Distribution
The expected distribution represents what you expect to observe in your frequency distribution based on certain assumptions or a theoretical model. For instance, if we are studying the roll of a fair six-sided die, we expect each number to appear an equal number of times in a large number of rolls. If each face of the die is equally likely, then the expected frequency for each face in 60 rolls is 10. Comparing this to the observed results helps us use the chi-square test to see if the die is fair.
Right-Tailed Test
A right-tailed test in the context of the chi-square goodness-of-fit test means that we focus on the upper tail of the chi-square distribution. Because the chi-square statistic can only be zero or positive, we are interested in whether the observed statistic is significantly larger than what our model predicts. Large values in the right tail of the chi-square distribution indicate a greater difference between observed and expected frequencies, suggesting that the observed distribution does not fit the expected distribution well.
Hypothesis Testing
In hypothesis testing, we start with a null hypothesis and an alternative hypothesis. For the chi-square goodness-of-fit test, our null hypothesis generally states that there is no significant difference between the observed and expected distributions. The alternative hypothesis suggests that there is a significant difference. By calculating the chi-square statistic and comparing it to a critical value from the chi-square distribution, we can decide whether to reject the null hypothesis. If our chi-square statistic is large and falls into the right tail (beyond the critical value), we reject the null hypothesis. This indicates that the observed distribution differs significantly from the expected distribution, leading us to accept the alternative hypothesis.

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Most popular questions from this chapter

Determine the expected counts for each outcome. $$ \begin{array}{lllll} \hline \boldsymbol{n}=\mathbf{5 0 0} & & & & \\ \hline p_{i} & 0.2 & 0.1 & 0.45 & 0.25 \\ \hline \text { Expected counts } & & & & \\ \hline \end{array} $$

According to the manufacturer of M\&Ms, \(13 \%\) of the plain M\&Ms in a bag should be brown, \(14 \%\) yellow, \(13 \%\) red, \(24 \%\) blue \(, 20 \%\) orange, and \(16 \%\) green. A student randomly selected a bag of plain M\&Ms. He counted the number of \(\mathrm{M} \& \mathrm{Ms}\) that were each color and obtained the results shown in the table. Test whether plain M\&Ms follow the distribution stated by M\&M/Mars at the \(\alpha=0.05\) level of significance. $$ \begin{array}{lc} \text { Color } & \text { Frequency } \\ \hline \text { Brown } & 57 \\ \hline \text { Yellow } & 64 \\ \hline \text { Red } & 54 \\ \hline \text { Blue } & 75 \\ \hline \text { Orange } & 86 \\ \hline \text { Green } & 64\\\ \hline \end{array} $$

The following table contains the number of successes and failures for three categories of a variable. $$ \begin{array}{lccc} & \text { Category } 1 & \text { Category } 2 & \text { Category } 3 \\ \hline \text { Success } & 76 & 84 & 69 \\ \hline \text { Failure } & 44 & 41 & 49 \\ \hline \end{array} $$ Test whether the proportions are equal for each category at the \(\alpha=0.01\) level of significance.

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}: p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=\frac{1}{4}\) \(H_{1}\) : At least one of the proportions is different from the others. $$ \begin{array}{lcccc} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \text { Observed } & 30 & 20 & 28 & 22 \\ \hline \text { Expected } & 25 & 25 & 25 & 25 \\ \hline \end{array} $$

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}:\) The random variable \(X\) is binomial with \(n=4, p=0.3\) \(H_{1}:\) The random variable \(X\) is not binomial with \(n=4\) \(p=0.3\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 260 & 400 & 280 & 50 & 10 \\ \hline \text { Expected } & 240.1 & 411.6 & 264.6 & 75.6 & 8.1 \\ \hline \end{array} $$

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