Questions: Are the following events independent or dependent? Event A: A mosquito lands on your arm. Event B: You have a mosquito bite.

Are the following events independent or dependent? Event A: A mosquito lands on your arm. Event B: You have a mosquito bite.
Transcript text: Are the following events independent or dependent? Event A: A mosquito lands on your arm. Event B: You have a mosquito bite.
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Solution

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Solution Steps

Step 1: Observed Frequencies

We have the following observed frequencies in the contingency table:

\[ \begin{array}{|c|c|c|} \hline & \text{Mosquito Bite (B)} & \text{No Mosquito Bite (Not B)} \\ \hline \text{Mosquito Lands (A)} & 30 & 10 \\ \hline \text{No Mosquito Lands (Not A)} & 5 & 55 \\ \hline \end{array} \]

Step 2: Expected Frequencies

The expected frequencies for each cell are calculated as follows:

  • For cell (1, 1): \[ E = \frac{R_1 \times C_1}{N} = \frac{40 \times 35}{100} = 14.0 \]

  • For cell (1, 2): \[ E = \frac{R_1 \times C_2}{N} = \frac{40 \times 65}{100} = 26.0 \]

  • For cell (2, 1): \[ E = \frac{R_2 \times C_1}{N} = \frac{60 \times 35}{100} = 21.0 \]

  • For cell (2, 2): \[ E = \frac{R_2 \times C_2}{N} = \frac{60 \times 65}{100} = 39.0 \]

Thus, the expected frequencies are:

\[ \begin{array}{|c|c|c|} \hline & \text{Mosquito Bite (B)} & \text{No Mosquito Bite (Not B)} \\ \hline \text{Mosquito Lands (A)} & 14.0 & 26.0 \\ \hline \text{No Mosquito Lands (Not A)} & 21.0 & 39.0 \\ \hline \end{array} \]

Step 3: Chi-Square Test Statistic

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

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

Calculating for each cell:

  • For cell (1, 1): \[ O = 30, \quad E = 14.0, \quad \frac{(30 - 14.0)^2}{14.0} = 18.2857 \]

  • For cell (1, 2): \[ O = 10, \quad E = 26.0, \quad \frac{(10 - 26.0)^2}{26.0} = 9.8462 \]

  • For cell (2, 1): \[ O = 5, \quad E = 21.0, \quad \frac{(5 - 21.0)^2}{21.0} = 12.1905 \]

  • For cell (2, 2): \[ O = 55, \quad E = 39.0, \quad \frac{(55 - 39.0)^2}{39.0} = 6.5641 \]

Summing these values gives:

\[ \chi^2 = 18.2857 + 9.8462 + 12.1905 + 6.5641 = 44.0018 \]

Step 4: Critical Value and P-Value

The critical value at \(\alpha = 0.05\) for a Chi-Square distribution with 1 degree of freedom is:

\[ \chi^2_{\alpha, df} = \chi^2_{(0.05, 1)} = 3.8415 \]

The p-value is calculated as:

\[ P = P(\chi^2 > 44.0018) = 0.0 \]

Step 5: Conclusion

Since the p-value \( (0.0) \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. Therefore, the events are dependent.

Final Answer

\(\boxed{\text{The events are dependent.}}\)

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