Questions: 1/4/25, 12.43 PM Week 7 - Homework Assignment (pages 4 of 11) Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in the table provided. a) State the null and alternative hypotheses. b) What is the test statistic? c) What can you conclude at the 5% significance level? 20-30 30-40 40-50 50-60 Private Practice 16 40 38 6 Hospital 8 44 59 39

Solution Steps

Let \( H_0 \) be the null hypothesis and \( H_1 \) be the alternative hypothesis.

• Null Hypothesis (\( H_0 \)): The distribution of working hours is the same for private practice doctors and hospital doctors.
• Alternative Hypothesis (\( H_1 \)): The distribution of working hours is different for private practice doctors and hospital doctors.

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 \( j \),
• \( N \) is the total number of observations.

The expected frequencies are as follows:

• For cell (1, 1): \( E = \frac{100 \times 24}{250} = 9.6 \)
• For cell (1, 2): \( E = \frac{100 \times 84}{250} = 33.6 \)
• For cell (1, 3): \( E = \frac{100 \times 97}{250} = 38.8 \)
• For cell (1, 4): \( E = \frac{100 \times 45}{250} = 18.0 \)
• For cell (2, 1): \( E = \frac{150 \times 24}{250} = 14.4 \)
• For cell (2, 2): \( E = \frac{150 \times 84}{250} = 50.4 \)
• For cell (2, 3): \( E = \frac{150 \times 97}{250} = 58.2 \)
• For cell (2, 4): \( E = \frac{150 \times 45}{250} = 27.0 \)

The expected frequencies matrix is:

\[ \begin{bmatrix} 9.6 & 33.6 & 38.8 & 18.0 \\ 14.4 & 50.4 & 58.2 & 27.0 \end{bmatrix} \]

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): \( O = 16, E = 9.6 \Rightarrow \frac{(16 - 9.6)^2}{9.6} = 4.2667 \)
• For cell (1, 2): \( O = 40, E = 33.6 \Rightarrow \frac{(40 - 33.6)^2}{33.6} = 1.2190 \)
• For cell (1, 3): \( O = 38, E = 38.8 \Rightarrow \frac{(38 - 38.8)^2}{38.8} = 0.0165 \)
• For cell (1, 4): \( O = 6, E = 18.0 \Rightarrow \frac{(6 - 18.0)^2}{18.0} = 8.0000 \)
• For cell (2, 1): \( O = 8, E = 14.4 \Rightarrow \frac{(8 - 14.4)^2}{14.4} = 2.8444 \)
• For cell (2, 2): \( O = 44, E = 50.4 \Rightarrow \frac{(44 - 50.4)^2}{50.4} = 0.8127 \)
• For cell (2, 3): \( O = 59, E = 58.2 \Rightarrow \frac{(59 - 58.2)^2}{58.2} = 0.0110 \)
• For cell (2, 4): \( O = 39, E = 27.0 \Rightarrow \frac{(39 - 27.0)^2}{27.0} = 5.3333 \)

Summing these values gives:

\[ \chi^2 = 4.2667 + 1.2190 + 0.0165 + 8.0000 + 2.8444 + 0.8127 + 0.0110 + 5.3333 = 22.5037 \]

The critical value for a Chi-Square distribution with \( df = 3 \) at \( \alpha = 0.05 \) is:

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

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

\[ P = P(\chi^2 > 22.5037) = 0.0001 \]

Final Answer