9. Mickey D's In a study of the accuracy of fast food drive-through orders, McDonald's had 33 orders that were not accurate among 362 orders observed (based on data from $$ Q S R $$ magazine). Use a 0.05 significance level to test the claim that the rate of inaccurate orders is equal to $$ 10 \% $$. Does the accuracy rate appear to be acceptable?

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To determine whether the rate of inaccurate McDonald's drive-through orders is acceptably low, we'll perform a hypothesis test for a population proportion. Here's a step-by-step analysis:

### **1. Define the Hypotheses**

- **Null Hypothesis ($$ H_0 $$)**: The rate of inaccurate orders is equal to $$ 10\% $$.
  
  $$
  H_0: p = 0.10
  $$

- **Alternative Hypothesis ($$ H_a $$)**: The rate of inaccurate orders is not equal to $$ 10\% $$.
  
  $$
  H_a: p \neq 0.10
  $$

*This is a two-tailed test since we're checking for any significant difference from $$ 10\% $$.*

### **2. Collect and Summarize the Data**

- **Sample Size ($$ n $$)**: 362 orders
- **Number of Inaccurate Orders ($$ X $$)**: 33
- **Sample Proportion ($$ \hat{p} $$)**:
  
  $$
  \hat{p} = \frac{X}{n} = \frac{33}{362} \approx 0.0912 \text{ or } 9.12\%
  $$

### **3. 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:
- $$ p_0 = 0.10 $$ (the hypothesized proportion)

Plugging in the values:

$$
z = \frac{0.0912 - 0.10}{\sqrt{\frac{0.10 \times 0.90}{362}}} = \frac{-0.0088}{\sqrt{\frac{0.09}{362}}} \approx \frac{-0.0088}{0.0158} \approx -0.559
$$

### **4. Determine the p-Value**

Using the standard normal distribution:

- **Find $$ P(Z > |z|) $$** when $$ z = -0.559 $$.

From standard normal tables or a calculator:

$$
P(Z > 0.559) \approx 0.2877
$$

Since it's a two-tailed test:

$$
\text{p-value} = 2 \times 0.2877 \approx 0.5754
$$

### **5. Make a Decision**

- **Significance Level ($$ \alpha $$)**: 0.05
- **Compare p-value with $$ \alpha $$**:
  
  $$
  0.5754 > 0.05
  $$

- **Decision**: **Fail to reject** the null hypothesis.

### **6. Conclusion**

There is **insufficient evidence** at the $$ 0.05 $$ significance level to conclude that the rate of inaccurate McDonald's drive-through orders differs from $$ 10\% $$.

**Interpretation for Acceptability**:
Since the observed inaccurate rate is approximately $$ 9.12\% $$, which is not significantly different from the claimed $$ 10\% $$, the accuracy rate **appears to be acceptable** based on this study.

### **Summary**

- **Test Used**: Two-tailed z-test for population proportion.
- **Test Statistic**: $$ z \approx -0.559 $$
- **p-Value**: $$ \approx 0.5754 $$
- **Conclusion**: Fail to reject $$ H_0 $$; the accuracy rate is acceptable.
There is not enough evidence to conclude that the rate of inaccurate McDonald's drive-through orders differs from 10%. Therefore, the accuracy rate appears to be acceptable.

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