Aliens from Planets X, Y, and Z gather on a remote planet every year to decide intergalactic policy. The organizing committee wants to check that the numbers of visitors from each planet is representative of that planet's population. Note that
\displaystyle \text {Population of Planet X: 1 million}
\displaystyle \text {Population of Planet Y: 4 million}
\displaystyle \text {Population of Planet Z: 5 million}.
Let E = \{ X, Y, Z\} denote the sample space. There are a total of 100 visitors chosen for this year's meeting from the overall population of 10 million. Let \xi _1, \ldots , \xi _{100} denote random variables corresponding to alien 1,2,\ldots , 100, respectively, so that
\xi _ i = \begin{cases} X \quad \text {if alien i comes from Planet X}\\ Y \quad \text {if alien i comes from Planet Y} \\ Z \quad \text {if alien i comes from Planet Z} \end{cases}
The organizing committee models the outcome of the selection process as a statistical experiment with a categorical distributional model: (\{ X, Y, Z\} , \{ \mathbf{P}_{\mathbf{p}} \} _{\mathbf{p} \in \Delta _3 }) and write \xi _1, \ldots , \xi _{100} \stackrel{iid}{\sim } \mathbf{P}_{\mathbf{p}^*} where \mathbf{p}^* is the true parameter. The null hypothesis and alternative hypothesis are, respectively,
\displaystyle H_0 \displaystyle : \mathbf{p}^* = \begin{bmatrix} \frac{1}{10} \\ \frac{4}{10} \\ \frac{5}{10} \end{bmatrix}.
\displaystyle H_1 \displaystyle : \mathbf{p}^* \neq \begin{bmatrix} \frac{1}{10} \\ \frac{4}{10} \\ \frac{5}{10} \end{bmatrix}.
Remark: Note that if H_0 holds, then the visiting delegation is representative of the populations of the three planets in the sense that the percentage of visitors from Planet X (respectively, Planet Y and Z) is not far from the percentage of aliens that live on Planet X (respectively, Planet Y and Z).
Suppose there are 20 visitors from Planet X, 30 visitors from Planet Y, and 50 visitors from Planet Z. Let \widehat{\mathbf{p} } denote the MLE for \mathbf{p}^* for this data set.
What is the asymptotic p-value of the \chi ^2 test
\psi _{100} = \mathbf{1}\left( 100 \left( \frac{(\widehat{p}_1 - \frac{1}{10} )^2}{1/10 } + \frac{(\widehat{p }_2 - \frac{4}{10} )^2}{4/10 } + \frac{(\widehat{p}_3 - \frac{5}{10} )^2}{5/10} \right) > C \right) ?
Use this tool to find the tail probabilities of a \chi ^2 distribution (you may also use any software you are familiar with). If you are using this tool, note that you need to set "Choose Type of Control" to "Adjust X-axis quantile (Chi square) value" to find the tail probability associated with an x-axis value for a chi-squared distribution with degrees of freedom set in the "Degrees of Freedom" box.
Give a numerical value with at least 5 decimals. (You are encouraged to also use computational tools.)
To find the asymptotic p-value for the chi-square test, we first need to compute the test statistic.
The test statistic can be calculated using the formula:
\(\psi _{100} = \mathbf{1}\left( 100 \left( \frac{(\widehat{p}_1 - \frac{1}{10} )^2}{1/10 } + \frac{(\widehat{p }_2 - \frac{4}{10} )^2}{4/10 } + \frac{(\widehat{p}_3 - \frac{5}{10} )^2}{5/10} \right) > C \right)\)
where \(\widehat{p}_1\), \(\widehat{p}_2\), and \(\widehat{p}_3\) are the estimated proportions of visitors from each planet, and C is a chi-square quantile value that determines the significance level of the test.
Given that there are 20 visitors from Planet X, 30 visitors from Planet Y, and 50 visitors from Planet Z, we can calculate the estimated proportions as follows:
\(\widehat{p}_1 = \frac{20}{100} = 0.2\)
\(\widehat{p}_2 = \frac{30}{100} = 0.3\)
\(\widehat{p}_3 = \frac{50}{100} = 0.5\)
Substituting these values into the test statistic formula, we have:
\(\psi _{100} = \mathbf{1}\left( 100 \left( \frac{(0.2 - \frac{1}{10} )^2}{1/10 } + \frac{(0.3 - \frac{4}{10} )^2}{4/10 } + \frac{(0.5 - \frac{5}{10} )^2}{5/10} \right) > C \right)\)
Simplifying the expression inside the parentheses:
\(\frac{(0.2 - \frac{1}{10} )^2}{1/10 } + \frac{(0.3 - \frac{4}{10} )^2}{4/10 } + \frac{(0.5 - \frac{5}{10} )^2}{5/10} = \frac{(0.1)^2}{1/10 } + \frac{(0.1)^2}{4/10 } + \frac{(0)^2}{5/10} = 0.01 + 0.025 + 0 = 0.035\)
Now, we need to determine the chi-square quantile value, C, that corresponds to the desired significance level. Since the problem does not specify a significance level, we will assume a standard significance level of 0.05 (95% confidence).
Using a chi-square distribution with 2 degrees of freedom (3 categories - 1), we can find the quantile value using a chi-square calculator or software. In this case, the quantile value is approximately 5.991.
Substituting these values into the test statistic, we have:
\(\psi _{100} = \mathbf{1}\left( 100 \cdot 0.035 > 5.991 \right)\)
Simplifying the inequality:
\(\psi _{100} = \mathbf{1}\left( 3.5 > 5.991 \right)\)
Since 3.5 is not greater than 5.991, the indicator function evaluates to 0. Therefore, the test statistic is false.
The p-value for the test can be calculated as the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, the p-value is equal to the tail probability of the chi-square distribution with 2 degrees of freedom greater than the test statistic value.
Using a chi-square distribution calculator or software, we can find the p-value for a chi-square statistic of 3.5 with 2 degrees of freedom. The p-value is approximately 0.17462.
So, the asymptotic p-value for the chi-square test is approximately 0.17462.
To find the asymptotic p-value of the chi-squared test, we need to calculate the test statistic and compare it to the critical value.
First, let's calculate the observed proportions or the estimates of the parameters:
\(\widehat{p}_1\) = Proportion of visitors from Planet X = 20/100 = 0.2
\(\widehat{p}_2\) = Proportion of visitors from Planet Y = 30/100 = 0.3
\(\widehat{p}_3\) = Proportion of visitors from Planet Z = 50/100 = 0.5
Next, we'll calculate the chi-squared test statistic using the formula:
\(\psi_{100} = 100 \left( \frac{(\widehat{p}_1 - \frac{1}{10})^2}{\frac{1}{10}} + \frac{(\widehat{p}_2 - \frac{4}{10})^2}{\frac{4}{10}} + \frac{(\widehat{p}_3 - \frac{5}{10})^2}{\frac{5}{10}} \right)\)
\(\psi_{100} = 100 \left( \frac{(0.2 - \frac{1}{10})^2}{\frac{1}{10}} + \frac{(0.3 - \frac{4}{10})^2}{\frac{4}{10}} + \frac{(0.5 - \frac{5}{10})^2}{\frac{5}{10}} \right)\)
\(\psi_{100} = 100 \left( \frac{(0.2 - 0.1)^2}{0.1} + \frac{(0.3 - 0.4)^2}{0.4} + \frac{(0.5 - 0.5)^2}{0.5} \right)\)
\(\psi_{100} = 100 \left( \frac{0.01}{0.1} + \frac{0.01}{0.4} + \frac{0}{0.5} \right)\)
\(\psi_{100} = 100 \left( 0.1 + 0.025 + 0 \right)\)
\(\psi_{100} = 12.5\)
Now, we need to find the critical value for the chi-square distribution with 2 degrees of freedom. We can use a chi-square distribution calculator or software to find the tail probability associated with an x-axis value of 12.5.
Using a chi-square distribution calculator (with 2 degrees of freedom), we find that the tail probability for an x-axis value of 12.5 is approximately 0.47618.
Finally, we can calculate the asymptotic p-value using the tail probability:
p-value = 2 * min (tail probability, 1 - tail probability)
p-value = 2 * min (0.47618, 1 - 0.47618)
p-value = 2 * min (0.47618, 0.52382)
p-value = 2 * 0.47618
p-value = 0.95236
Therefore, the asymptotic p-value of the chi-squared test is approximately 0.95236.