Question 1162408
```python?code_reference&code_event_index=3
import numpy as np
from scipy import stats

data = [
    42, 39, 42, 45, 43, 40, 39, 41, 40, 42,
    43, 42, 45, 47, 38, 46, 48, 42, 47, 41,
    40, 44, 43, 42, 40, 41, 42, 39, 40, 41
]

mu_0 = 43
n = len(data)
x_bar = np.mean(data)
s = np.std(data, ddof=1)
se = s / np.sqrt(n)
t_stat = (x_bar - mu_0) / se
df = n - 1

# One-tailed test (Reduced mean: HA: mu < 43)
p_value = stats.t.cdf(t_stat, df)

print(f"{x_bar=}")
print(f"{s=}")
print(f"{t_stat=}")
print(f"{p_value=}")

```
```text?code_stdout&code_event_index=3
x_bar=42.13333333333333
s=2.582879085016581
t_stat=-1.8378440010020713
p_value=0.038175278168928364

```
To determine if the mean length of the bars has reduced, we conduct a **one-sample t-test**.

### 1. State the Hypotheses
* **Null Hypothesis ($H_0$):** $\mu \geq 43$ (The mean length has not reduced)
* **Alternative Hypothesis ($H_A$):** $\mu < 43$ (The mean length has reduced)

### 2. Identify Sample Statistics
From the provided data of $n = 30$ bars:
* **Sample Mean ($\bar{x}$):** $42.13$ mm
* **Sample Standard Deviation ($s$):** $2.58$ mm
* **Hypothesized Mean ($\mu_0$):** $43$ mm

### 3. Calculate the Test Statistic ($t$)
The $t$-statistic measures how many standard errors the sample mean is from the hypothesized mean:
$$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{42.13 - 43}{2.58 / \sqrt{30}} \approx -1.838$$

### 4. Determine the p-value
Using the $t$-distribution with $df = n - 1 = 29$:
* **$p$-value:** $\approx 0.0382$

### 5. Conclusion
At a standard significance level of **$\alpha = 0.05$**:
* Since the $p$-value ($0.0382$) is **less than** $0.05$, we **reject the null hypothesis**.

**Final Result:** Yes, it is **reasonable to conclude** that the mean length of the bars has reduced. The statistical evidence indicates that the decrease from $43$ mm to $42.13$ mm is significant enough that it is unlikely to have occurred by random chance alone. 

> **Note:** If a more stringent significance level like $\alpha = 0.01$ were used, we would fail to reject the null hypothesis, as $0.0382 > 0.01$. However, in most industrial engineering contexts, $0.05$ is the standard threshold for investigation.