To determine if the mean number of diners changed with the free dessert offer, we will conduct a hypothesis test for the mean using the P-value method at $\alpha = 0.01$. Below are the steps:

---

Step 1: Formulate Hypotheses

• Null hypothesis ($H_0$): The mean number of diners did not change, i.e., $\mu = 150$.
• Alternative hypothesis ($H_a$): The mean number of diners changed, i.e., $\mu \neq 150$.

---

Step 2: Gather Sample Data

The sample data consists of the following observations: $185, 148, 170, 131, 191, 118, 121, 130, 153, 189, 171, 131$

• Sample size ($n$) = 12
• Sample mean ($\bar{x}$) = $\frac{\text{sum of data}}{n}$
• Sample standard deviation ($s$) = $\sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}$

---

Step 3: Compute Test Statistic

The test statistic for a single mean is calculated as: $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ where:

• $\mu_0 = 150$ (mean under $H_0$),
• $s$ is the sample standard deviation,
• $n$ is the sample size.

---

Step 4: Determine P-value

Using the $t$-distribution with $n - 1 = 11$ degrees of freedom, we will calculate the two-tailed $P$-value.

---

Step 5: Compare P-value with $\alpha$

If $P \leq \alpha$, reject $H_0$. Otherwise, fail to reject $H_0$.

---

Let’s calculate the necessary values.import numpy as np import scipy.stats as stats

Data

data = np.array([185, 148, 170, 131, 191, 118, 121, 130, 153, 189, 171, 131]) mu_0 = 150 # Null hypothesis mean alpha = 0.01 # Significance level

Calculations

n = len(data) sample_mean = np.mean(data) sample_std = np.std(data, ddof=1) t_statistic = (sample_mean - mu_0) / (sample_std / np.sqrt(n))

P-value (two-tailed test)

p_value = 2 * stats.t.sf(np.abs(t_statistic), df=n-1)

Display results

sample_mean, sample_std, t_statistic, p_value(153.16666666666666, 27.21574856029078, 0.40306277412513153, 0.6946253906548336)(153.16666666666666, 27.21574856029078, 0.40306277412513153, 0.6946253906548336)success### Results:

• Sample mean ($\bar{x}$) = 153.17
• Sample standard deviation ($s$) = 27.22
• Test statistic ($t$) = 0.403
• $P$-value = 0.695

---

Step 6: Decision

Since the $P$-value (0.695) is greater than the significance level ($\alpha = 0.01$), we fail to reject the null hypothesis.

Conclusion

There is not enough evidence to conclude that the mean number of diners changed while the free dessert offer was in effect.

---

Do you have further questions or need detailed step-by-step explanations? Here are 5 related questions:

1. How is the test statistic $t$ calculated in this problem?
2. What does it mean to fail to reject the null hypothesis in practical terms?
3. Can we perform a similar test if the population standard deviation is known?
4. What are the assumptions of a $t$-test, and are they satisfied here?
5. How does the significance level ($\alpha$) affect the conclusion?

Tip: Always verify assumptions, such as normality of data, when conducting a $t$-test!