8) A local supermarket spent $$ \$ 250,000 $$ remodeling. Prior to remodeling, the average daily receipts were $$ \$ 57,500 $$. In a sample of 18 days following the remodeling, daily receipts were $$ \$ 59,800 $$ with a standard deviation of $$ \$ 5,460 $$. At the $$ 10 \% $$ significance level, what can you conclude as to whether the remodeling improved daily receipts?

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UpStudy AI · Accepted Answer

To determine whether the remodeling of the supermarket led to an improvement in daily receipts, we'll perform a hypothesis test using the provided data.

### **Given:**
- **Prior to Remodeling:**
  - Mean daily receipts ($$ \mu_0 $$) = \$57,500
- **After Remodeling (Sample):**
  - Sample size ($$ n $$) = 18 days
  - Sample mean ($$ \bar{x} $$) = \$59,800
  - Sample standard deviation ($$ s $$) = \$5,460
- **Significance Level ($$ \alpha $$)** = 10% (0.10)

### **Hypotheses:**
- **Null Hypothesis ($$ H_0 $$)**: The remodeling did **not** improve daily receipts.
  $$
  H_0: \mu \leq 57,500
  $$
- **Alternative Hypothesis ($$ H_1 $$)**: The remodeling **did** improve daily receipts.
  $$
  H_1: \mu > 57,500
  $$

### **Test Statistic:**
Since the population standard deviation is unknown and the sample size is small ($$ n < 30 $$), we'll use the **t-test**.

$$
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
$$

Plugging in the values:

$$
t = \frac{59,800 - 57,500}{5,460 / \sqrt{18}} = \frac{2,300}{1,285.93} \approx 1.787
$$

### **Critical Value:**
For a one-tailed test at $$ \alpha = 0.10 $$ and degrees of freedom ($$ df = n - 1 = 17 $$), the critical t-value ($$ t_{\text{critical}} $$) is approximately **1.333**.

### **Decision Rule:**
- **If** $$ t > t_{\text{critical}} $$, **reject** $$ H_0 $$.
- **Else**, **do not reject** $$ H_0 $$.

### **Conclusion:**
Since $$ t \approx 1.787 $$ **is greater** than $$ t_{\text{critical}} \approx 1.333 $$:

- **Reject the null hypothesis**.

**Interpretation:**  
At the 10% significance level, there is sufficient evidence to conclude that the remodeling **improved** the supermarket's daily receipts.

**Final Answer:**
At the 10% level, the test rejects H₀. There is sufficient evidence to conclude remodeling increased daily receipts.
At the 10% significance level, the test rejects the null hypothesis. There is sufficient evidence to conclude that the remodeling increased daily receipts.

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