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

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
Transcript text: 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? \begin{tabular}{|c|c|c|c|} \hline \multirow{4}{*}{ Seed Type } & & \multicolumn{2}{|c|}{ Observed Frequencies } \\ \cline { 2 - 4 } & & Germinated & Failed to Germinate \\ \cline { 2 - 4 } & 1 & 39 & 9 \\ \cline { 2 - 4 } & 2 & 54 & 34 \\ \cline { 2 - 4 } & 3 & 88 & 63 \\ \cline { 2 - 4 } & 4 & 57 & 42 \\ \hline \end{tabular} What is the Expected Count for Seed Type = 1 and a Failed to Germinated Seed? 18.404 33.741 57.896 37.959
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Solution

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Solution Steps

Step 1: Observed Frequencies

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

Seed TypeGerminatedFailed to Germinate1399254343886345742 \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}

Step 2: Expected Frequencies

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

E=Ri×CjN E = \frac{R_i \times C_j}{N}

where RiR_i is the total for row ii, CjC_j is the total for column jj, and NN is the grand total.

The expected frequencies are:

Seed TypeGerminatedFailed to Germinate129.595918.4041254.259133.7409393.103657.8964461.041537.9585 \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}

Step 3: Chi-Square Test Statistic

The Chi-Square test statistic (χ2\chi^2) is calculated using the formula:

χ2=(OE)2E \chi^2 = \sum \frac{(O - E)^2}{E}

where OO is the observed frequency and EE is the expected frequency. The calculations for each cell are as follows:

  • For cell (1, 1): O=39,E=29.5959,(3929.5959)229.5959=2.9882O = 39, E = 29.5959, \frac{(39 - 29.5959)^2}{29.5959} = 2.9882
  • For cell (1, 2): O=9,E=18.4041,(918.4041)218.4041=4.8053O = 9, E = 18.4041, \frac{(9 - 18.4041)^2}{18.4041} = 4.8053
  • For cell (2, 1): O=54,E=54.2591,(5454.2591)254.2591=0.0012O = 54, E = 54.2591, \frac{(54 - 54.2591)^2}{54.2591} = 0.0012
  • For cell (2, 2): O=34,E=33.7409,(3433.7409)233.7409=0.0020O = 34, E = 33.7409, \frac{(34 - 33.7409)^2}{33.7409} = 0.0020
  • For cell (3, 1): O=88,E=93.1036,(8893.1036)293.1036=0.2798O = 88, E = 93.1036, \frac{(88 - 93.1036)^2}{93.1036} = 0.2798
  • For cell (3, 2): O=63,E=57.8964,(6357.8964)257.8964=0.4499O = 63, E = 57.8964, \frac{(63 - 57.8964)^2}{57.8964} = 0.4499
  • For cell (4, 1): O=57,E=61.0415,(5761.0415)261.0415=0.2676O = 57, E = 61.0415, \frac{(57 - 61.0415)^2}{61.0415} = 0.2676
  • For cell (4, 2): O=42,E=37.9585,(4237.9585)237.9585=0.4303O = 42, E = 37.9585, \frac{(42 - 37.9585)^2}{37.9585} = 0.4303

Summing these values gives:

χ2=9.2243 \chi^2 = 9.2243

Step 4: Degrees of Freedom and Critical Value

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

df=(r1)(c1)=(41)(21)=3 df = (r - 1)(c - 1) = (4 - 1)(2 - 1) = 3

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

χα,df2=7.8147 \chi^2_{\alpha, df} = 7.8147

Step 5: P-Value

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

P=P(χ2>9.2243)=0.0265 P = P(\chi^2 > 9.2243) = 0.0265

Final Answer

Since the calculated Chi-Square statistic 9.22439.2243 is greater than the critical value 7.81477.8147 and the p-value 0.02650.0265 is less than 0.050.05, we reject the null hypothesis. Thus, we conclude that the proportion of seeds that germinate is dependent on the seed type.

18.4041 \boxed{18.4041}

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