Questions: Question 9 (1 point) The data presented in the table below resulted from an experiment in which seeds of 4 different types were planted and the number of seeds that germinated within 4 weeks after planting was recorded for each seed type. At the .05 level of significance, is the proportion of seeds that germinate dependent on the seed type? Seed Type Germinated Failed to Germinate 1 39 9 2 54 34 3 88 63 4 57 42 What is the Expected Count for Seed Type = 1 and a Failed to Germinated Seed? - 18.404 - 33.741 - 57.896 - 37.959

Solution Steps

The observed frequencies for the germination of seeds are as follows:

\[ \begin{array}{|c|c|c|} \hline \text{Seed Type} & \text{Germinated} & \text{Failed to Germinate} \\ \hline 1 & 39 & 9 \\ 2 & 54 & 34 \\ 3 & 88 & 63 \\ 4 & 57 & 42 \\ \hline \end{array} \]

The expected frequencies for each cell in the contingency table are calculated using the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \(R_i\) is the total for row \(i\), \(C_j\) is the total for column \(j\), and \(N\) is the grand total.

The expected frequencies are:

\[ \begin{array}{|c|c|c|} \hline \text{Seed Type} & \text{Germinated} & \text{Failed to Germinate} \\ \hline 1 & 29.5959 & 18.4041 \\ 2 & 54.2591 & 33.7409 \\ 3 & 93.1036 & 57.8964 \\ 4 & 61.0415 & 37.9585 \\ \hline \end{array} \]

The Chi-Square test statistic (\(\chi^2\)) is calculated using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \(O\) is the observed frequency and \(E\) is the expected frequency. The calculations for each cell are as follows:

• For cell (1, 1): \(O = 39, E = 29.5959, \frac{(39 - 29.5959)^2}{29.5959} = 2.9882\)
• For cell (1, 2): \(O = 9, E = 18.4041, \frac{(9 - 18.4041)^2}{18.4041} = 4.8053\)
• For cell (2, 1): \(O = 54, E = 54.2591, \frac{(54 - 54.2591)^2}{54.2591} = 0.0012\)
• For cell (2, 2): \(O = 34, E = 33.7409, \frac{(34 - 33.7409)^2}{33.7409} = 0.0020\)
• For cell (3, 1): \(O = 88, E = 93.1036, \frac{(88 - 93.1036)^2}{93.1036} = 0.2798\)
• For cell (3, 2): \(O = 63, E = 57.8964, \frac{(63 - 57.8964)^2}{57.8964} = 0.4499\)
• For cell (4, 1): \(O = 57, E = 61.0415, \frac{(57 - 61.0415)^2}{61.0415} = 0.2676\)
• For cell (4, 2): \(O = 42, E = 37.9585, \frac{(42 - 37.9585)^2}{37.9585} = 0.4303\)

Summing these values gives:

\[ \chi^2 = 9.2243 \]

The degrees of freedom (\(df\)) for this test is calculated as:

\[ df = (r - 1)(c - 1) = (4 - 1)(2 - 1) = 3 \]

The critical value at \(\alpha = 0.05\) for \(df = 3\) is:

\[ \chi^2_{\alpha, df} = 7.8147 \]

The p-value associated with the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 9.2243) = 0.0265 \]

Final Answer