Questions: (a) When α=0.20 and n=17, xleft^2=9.31222 xright^2=23.5418 (b) When α=0.05 and n=31, xtet^2=16.7908 xright^2=46.9792 (c) When α=0.10 and n=25, xtett^2= xtight^2=

Solution Steps

We are tasked with calculating the Chi-Square test statistic and the critical value for a Chi-Square test with a significance level of \(\alpha = 0.10\) and \(n = 25\). The observed frequencies \(O_i\) and expected frequencies \(E_i\) are provided as follows:

• Observed frequencies: \(O = [5, 10, 10]\)
• Expected frequencies: \(E = [8, 8, 9]\)

The Chi-Square test statistic \(\chi^2\) is calculated using the formula: \[ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} \] Substituting the observed and expected values, we find: \[ \chi^2 = \frac{(5 - 8)^2}{8} + \frac{(10 - 8)^2}{8} + \frac{(10 - 9)^2}{9} = 1.7361 \]

The degrees of freedom \(df\) for the Chi-Square test is calculated as: \[ df = k - 1 \] where \(k\) is the number of categories. In this case, \(k = 3\), so: \[ df = 3 - 1 = 2 \]

The critical value for the Chi-Square test at \(\alpha = 0.10\) with \(df = 2\) is determined using the Chi-Square distribution: \[ \chi^2(1 - \alpha, df) = \chi^2(0.9, 2) = 4.6052 \]

The p-value associated with the Chi-Square test statistic is calculated as: \[ P = P(\chi^2 > 1.7361) = 0.4198 \]

The results of the calculations are as follows:

• Chi-Square test statistic: \(x_{\text{tett}}^2 = 1.7361\)
• Critical value: \(x_{\text{tight}}^2 = 4.6052\)

Final Answer