Question 1188762: 7. Fei Hung works in a noodle bar in Tsim Sha Tsui. She records the daily number of bowls of noodles she sells over the course of one particular week. The information is recorded in the table below. Day Mon Tues Wed Thurs Fri Sat Sun No. of bowls of 48 59 44 55 62 67 50 noodles sold. Perform a goodness of fit test at the 5% significance level to determine if Fei Hung's data fits a uniform distribution. You should state your null and alternative hypotheses and justify any conclusions found. The critical value for the test is 12.59. (7)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to perform a goodness-of-fit test to determine if Fei Hung's noodle sales data fits a uniform distribution:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The number of bowls of noodles sold each day follows a uniform distribution.
* **Alternative Hypothesis (H1):** The number of bowls of noodles sold each day does *not* follow a uniform distribution.
**2. Calculate Expected Frequencies:**
If the distribution is uniform, we expect the same number of bowls sold each day. Calculate the average number of bowls sold over the week:
(48 + 59 + 44 + 55 + 62 + 67 + 50) / 7 = 55
So, the expected frequency for each day is 55.
**3. Calculate the Chi-Square Statistic:**
The chi-square statistic measures the difference between the observed and expected frequencies. The formula is:
χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]
| Day | Observed (O) | Expected (E) | (O - E)² | (O - E)² / E |
|---------|--------------|--------------|----------|---------------|
| Mon | 48 | 55 | 49 | 0.891 |
| Tues | 59 | 55 | 16 | 0.291 |
| Wed | 44 | 55 | 121 | 2.2 |
| Thurs | 55 | 55 | 0 | 0 |
| Fri | 62 | 55 | 49 | 0.891 |
| Sat | 67 | 55 | 144 | 2.618 |
| Sun | 50 | 55 | 25 | 0.454 |
| **Total** | | | | **7.345** |
χ² ≈ 7.345
**4. Degrees of Freedom:**
Degrees of freedom (df) = Number of categories - 1 = 7 - 1 = 6
**5. Compare to Critical Value:**
The critical value given is 12.59 at a 5% significance level. Our calculated χ² (7.345) is *less* than the critical value (12.59).
**6. Conclusion:**
Since the calculated chi-square statistic is less than the critical value, we *fail to reject* the null hypothesis. There is not enough evidence at the 5% significance level to conclude that the number of bowls of noodles sold each day does *not* follow a uniform distribution. In other words, the data is consistent with a uniform distribution.
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