We will conduct a one-sample t-test to determine if the mean number of patients per day is significantly more than 25. The null hypothesis ($H_0$) and alternative hypothesis ($H_a$) are:

• $H_0$: The mean number of patients per day is $\mu = 25$.
• $H_a$: The mean number of patients per day is $\mu > 25$ (one-tailed test).

Steps:

1. Sample Data: $25, 27, 25, 26, 25, 28, 28, 27, 24, 26, 25, 29, 25, 27, 24$

   • Sample size ($n$) = 15

2. Calculate the Sample Mean ($\bar{x}$): $\bar{x} = \frac{\text{Sum of all values}}{n}$

3. Calculate the Sample Standard Deviation ($s$): $s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}}$

4. Test Statistic ($t$): $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$

5. Critical Value and p-value:

   • For $\alpha = 0.01$, find the critical value from the t-distribution with $n-1$ degrees of freedom ($df = 15-1=14$).
   • Compute the p-value.

Let’s calculate these values step by step.import numpy as np from scipy.stats import t

Sample data

data = [25, 27, 25, 26, 25, 28, 28, 27, 24, 26, 25, 29, 25, 27, 24] n = len(data) mu = 25 # Null hypothesis mean

Calculate sample mean

sample_mean = np.mean(data)

Calculate sample standard deviation

sample_std = np.std(data, ddof=1) # ddof=1 for sample standard deviation

Calculate t-statistic

t_statistic = (sample_mean - mu) / (sample_std / np.sqrt(n))

Degrees of freedom

df = n - 1

Calculate p-value for one-tailed test

p_value = 1 - t.cdf(t_statistic, df)

Critical t-value for alpha = 0.01

alpha = 0.01 critical_t = t.ppf(1 - alpha, df)

sample_mean, sample_std, t_statistic, critical_t, p_value(26.066666666666666, 1.533747356112131, 2.6935219933657377, 2.624494067560231, 0.008738459784443542)(26.066666666666666, 1.533747356112131, 2.6935219933657377, 2.624494067560231, 0.008738459784443542)success### Results:

1. Sample Mean ($\bar{x}$): $26.07$
2. Sample Standard Deviation ($s$): $1.53$
3. Test Statistic ($t$): $2.69$
4. Critical Value ($t_{critical}$): $2.62$ at $\alpha = 0.01$ and $df = 14$.
5. p-value: $0.0087$

Conclusion:

• Since $t = 2.69 > t_{critical} = 2.62$, we reject the null hypothesis ($H_0$).
• The p-value ($0.0087$) is less than the significance level ($\alpha = 0.01$), indicating strong evidence that the mean number of patients per day is greater than 25.

Interpretation:

The enlargement of the recovery room appears to have significantly increased the mean number of patients per day, with a p-value of $0.0087$. This suggests that there is less than a 1% probability that this result is due to random chance under the null hypothesis.

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Do you want further details on interpreting the t-test or visualizing the data?
Here are 5 related questions for exploration:

1. How does sample size affect the t-test results?
2. What is the difference between a one-tailed and two-tailed t-test?
3. How is the critical value determined for different significance levels?
4. What does the p-value indicate in hypothesis testing?
5. Could we use a different test for similar data?

Tip: Always ensure that the sample data meets the assumptions of the t-test, including normality and independence.