To determine if this year's distribution of patient wait times significantly differs from last year's, Averi can perform a Chi-Square Goodness-of-Fit test. Here is the step-by-step solution for this problem:
1. State the hypotheses:
- Null hypothesis (H₀): The current year's wait time distribution is the same as last year's distribution.
- Alternative hypothesis (H₁): The current year's wait time distribution is different from last year's distribution.
2. Observed frequencies:
The observed frequencies (from this year's sample) are:
- 0-15 minutes: 75 patients
- 16-30 minutes: 55 patients
- 31-45 minutes: 15 patients
- 46+ minutes: 5 patients
3. Expected frequencies:
The expected percentages (from last year's distribution) are:
- 0-15 minutes: 50%
- 16-30 minutes: 30%
- 31-45 minutes: 10%
- 46+ minutes: 10%
The total sample size is 150 patients. Using this, we can calculate the expected frequencies:
- For 0-15 minutes: [tex]\( 0.50 \times 150 = 75 \)[/tex]
- For 16-30 minutes: [tex]\( 0.30 \times 150 = 45 \)[/tex]
- For 31-45 minutes: [tex]\( 0.10 \times 150 = 15 \)[/tex]
- For 46+ minutes: [tex]\( 0.10 \times 150 = 15 \)[/tex]
4. Calculate the Chi-Square statistic:
The formula for the Chi-Square statistic is:
[tex]\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\][/tex]
where [tex]\(O_i\)[/tex] are the observed frequencies and [tex]\(E_i\)[/tex] are the expected frequencies.
Substituting in the values:
[tex]\[
\chi^2 = \frac{(75 - 75)^2}{75} + \frac{(55 - 45)^2}{45} + \frac{(15 - 15)^2}{15} + \frac{(5 - 15)^2}{15}
\][/tex]
[tex]\[
\chi^2 = \frac{0}{75} + \frac{100}{45} + \frac{0}{15} + \frac{100}{15}
\][/tex]
[tex]\[
\chi^2 = 0 + 2.2222 + 0 + 6.6667
\][/tex]
[tex]\[
\chi^2 \approx 8.8889
\][/tex]
5. Degrees of freedom:
The degrees of freedom (df) for a Chi-Square Goodness-of-Fit test is calculated as:
[tex]\[
\text{df} = \text{number of categories} - 1
\][/tex]
[tex]\[
\text{df} = 4 - 1 = 3
\][/tex]
6. Determine the p-value:
Using the Chi-Square distribution table or a statistical software, we can find the p-value corresponding to the Chi-Square statistic of approximately 8.8889 with 3 degrees of freedom. The p-value is approximately 0.0308.
7. Decision:
- If the p-value is less than the chosen significance level (typically 0.05), we reject the null hypothesis. In this case, the p-value is 0.0308, which is less than 0.05.
- Therefore, we reject the null hypothesis.
8. Conclusion:
There is sufficient evidence to conclude that the distribution of patient wait times this year is significantly different from last year's distribution.
