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

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
Transcript text: 1/4/25, 12.43 PM Week 7 - Homework Assigniment (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? \begin{tabular}{|l|l|l|l|l|} \hline & $20-30$ & $30-40$ & $\mathbf{4 0 - 5 0}$ & $50-60$ \\ \hline Private Practice & 16 & 40 & 38 & 6 \\ \hline Hospital & 8 & 44 & 59 & 39 \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: State the Null and Alternative Hypotheses

Let H0 H_0 be the null hypothesis and H1 H_1 be the alternative hypothesis.

  • Null Hypothesis (H0 H_0 ): The distribution of working hours is the same for private practice doctors and hospital doctors.
  • Alternative Hypothesis (H1 H_1 ): The distribution of working hours is different for private practice doctors and hospital doctors.
Step 2: Calculate the 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:

  • Ri R_i is the total for row i i ,
  • Cj C_j is the total for column j j ,
  • N N is the total number of observations.

The expected frequencies are as follows:

  • For cell (1, 1): E=100×24250=9.6 E = \frac{100 \times 24}{250} = 9.6
  • For cell (1, 2): E=100×84250=33.6 E = \frac{100 \times 84}{250} = 33.6
  • For cell (1, 3): E=100×97250=38.8 E = \frac{100 \times 97}{250} = 38.8
  • For cell (1, 4): E=100×45250=18.0 E = \frac{100 \times 45}{250} = 18.0
  • For cell (2, 1): E=150×24250=14.4 E = \frac{150 \times 24}{250} = 14.4
  • For cell (2, 2): E=150×84250=50.4 E = \frac{150 \times 84}{250} = 50.4
  • For cell (2, 3): E=150×97250=58.2 E = \frac{150 \times 97}{250} = 58.2
  • For cell (2, 4): E=150×45250=27.0 E = \frac{150 \times 45}{250} = 27.0

The expected frequencies matrix is:

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

Step 3: Calculate the 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 O O is the observed frequency and E E is the expected frequency. The calculations for each cell are as follows:

  • For cell (1, 1): O=16,E=9.6(169.6)29.6=4.2667 O = 16, E = 9.6 \Rightarrow \frac{(16 - 9.6)^2}{9.6} = 4.2667
  • For cell (1, 2): O=40,E=33.6(4033.6)233.6=1.2190 O = 40, E = 33.6 \Rightarrow \frac{(40 - 33.6)^2}{33.6} = 1.2190
  • For cell (1, 3): O=38,E=38.8(3838.8)238.8=0.0165 O = 38, E = 38.8 \Rightarrow \frac{(38 - 38.8)^2}{38.8} = 0.0165
  • For cell (1, 4): O=6,E=18.0(618.0)218.0=8.0000 O = 6, E = 18.0 \Rightarrow \frac{(6 - 18.0)^2}{18.0} = 8.0000
  • For cell (2, 1): O=8,E=14.4(814.4)214.4=2.8444 O = 8, E = 14.4 \Rightarrow \frac{(8 - 14.4)^2}{14.4} = 2.8444
  • For cell (2, 2): O=44,E=50.4(4450.4)250.4=0.8127 O = 44, E = 50.4 \Rightarrow \frac{(44 - 50.4)^2}{50.4} = 0.8127
  • For cell (2, 3): O=59,E=58.2(5958.2)258.2=0.0110 O = 59, E = 58.2 \Rightarrow \frac{(59 - 58.2)^2}{58.2} = 0.0110
  • For cell (2, 4): O=39,E=27.0(3927.0)227.0=5.3333 O = 39, E = 27.0 \Rightarrow \frac{(39 - 27.0)^2}{27.0} = 5.3333

Summing these values gives:

χ2=4.2667+1.2190+0.0165+8.0000+2.8444+0.8127+0.0110+5.3333=22.5037 \chi^2 = 4.2667 + 1.2190 + 0.0165 + 8.0000 + 2.8444 + 0.8127 + 0.0110 + 5.3333 = 22.5037

Step 4: Determine the Critical Value and P-Value

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

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

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

P=P(χ2>22.5037)=0.0001 P = P(\chi^2 > 22.5037) = 0.0001

Final Answer

Since the calculated Chi-Square statistic 22.5037 22.5037 is greater than the critical value 7.8147 7.8147 and the p-value 0.0001 0.0001 is less than α=0.05 \alpha = 0.05 , we reject the null hypothesis.

Reject H0: There is a significant difference in the distribution of working hours between private practice and hospital doctors. \boxed{\text{Reject } H_0: \text{ There is a significant difference in the distribution of working hours between private practice and hospital doctors.}}

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