SOLUTION: In a study of the accuracy of fast food drive-through orders, one restaurant had 31 orders that were not accurate among 356 orders observed. Use a 0.01 significance level to tes
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Question 1168833: In a study of the accuracy of fast food drive-through orders, one restaurant had 31 orders that were not accurate among 356 orders observed. Use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?
Need to find
Hupothesis test
p-value
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this hypothesis test step-by-step.
**1. Define the Hypotheses**
* **Null Hypothesis (H₀):** The proportion of inaccurate orders is equal to 10% (p = 0.10).
* **Alternative Hypothesis (H₁):** The proportion of inaccurate orders is not equal to 10% (p ≠ 0.10).
This is a two-tailed test.
**2. Set the Significance Level**
* α = 0.01
**3. Calculate the Sample Proportion**
* Sample size (n) = 356
* Number of inaccurate orders (x) = 31
* Sample proportion (p̂) = x/n = 31/356 ≈ 0.08708
**4. Calculate the Test Statistic**
We'll use the z-test for proportions:
$$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$
Where:
* p̂ = sample proportion (0.08708)
* p₀ = hypothesized proportion (0.10)
* n = sample size (356)
$$ z = \frac{0.08708 - 0.10}{\sqrt{\frac{0.10(1-0.10)}{356}}} $$
$$ z = \frac{-0.01292}{\sqrt{\frac{0.09}{356}}} $$
$$ z = \frac{-0.01292}{\sqrt{0.000252808}} $$
$$ z = \frac{-0.01292}{0.0158999} $$
$$ z \approx -0.8126 $$
**5. Calculate the P-value**
Since this is a two-tailed test, we need to find the area in both tails beyond z = -0.8126 and z = 0.8126.
* P(Z < -0.8126) ≈ 0.2081
* P(Z > 0.8126) ≈ 0.2081
* P-value = 2 * 0.2081 ≈ 0.4162
**6. Make a Decision**
* Compare the p-value (0.4162) with the significance level (0.01).
* Since 0.4162 > 0.01, we fail to reject the null hypothesis.
**7. Conclusion**
There is not sufficient evidence at the 0.01 significance level to reject the claim that the rate of inaccurate orders is equal to 10%.
**Does the accuracy rate appear to be acceptable?**
The sample proportion of inaccurate orders (approximately 8.71%) is slightly lower than the claimed 10%. Statistically, we can't reject the 10% rate. Whether the accuracy rate is "acceptable" is a subjective judgment.
* From the statistical test, there is no evidence that it is different than 10%.
* Whether it is acceptable is up to the restaurant, and the customers. 8.7% is still a fairly high error rate.
**Summary**
* **Hypothesis test:** Two-tailed z-test for proportions.
* **p-value:** approximately 0.4162.
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