Example 2: The head of the faculty library states that the average number of students who visit it per day is 350. To confirm or not this assumption, the number of attendees for 30 days is recorded and it is found that the average was 372.6. If the standard deviation of 52.414 and a significance level of 0.05 is considered, say whether you would support what the head of the library says.

We can use a one-sample t-test to determine if the average number of students who visit the library per day is significantly different from 350.

Given data:
- Null hypothesis, Ho: \mu = 350 (average number of students per day is 350)
- Alternative hypothesis, Ha: \mu \neq 350 (average number of students per day is not 350)
- Sample mean, \bar{x} = 372.6
- Population standard deviation, \sigma = 52.414
- Sample size, n = 30
- Significance level, \alpha = 0.05

Calculate the t-score:
t = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

Substitute the values:
t = \frac{372.6 - 350}{\frac{52.414}{\sqrt{30}}}
t\approx2.362

Find the critical t-value at a significance level of 0.05 with degrees of freedom (df) = n-1 = 29 using a t-table or calculator.

Since this is a two-tailed test, we need to look for the critical t-value at \alpha/2 = 0.025 for df = 29.

The critical t-value is approximately \pm 2.045 for df = 29.

Compare the calculated t-value with the critical t-value:
Since 2.366 > 2.045, we reject the null hypothesis.

Therefore, we have enough evidence to reject the head of the faculty library's claim that the average number of students who visit the library per day is 350. Instead, the average number of students is significantly different from 350.

\boxed{ \text{Answer: We reject the null hypothesis. The head of the faculty library claim is not supported.} }