Let's analyze and solve the question step by step.
Given data:
1. Sample Mean: [tex]\( \bar{x} = 3.158191551 \)[/tex]
2. Sample Variance: [tex]\( s^2 = 1.248743026 \)[/tex]
3. Number of Observations: [tex]\( n = 300 \)[/tex]
4. Hypothesized Mean: [tex]\( \mu_0 = 3 \)[/tex]
5. Degrees of Freedom: [tex]\( df = 299 \)[/tex] (since [tex]\( df = n - 1 \)[/tex])
6. t Stat: [tex]\( t = 2.451926089 \)[/tex]
7. [tex]\( P ( T \leq t ) \)[/tex] one-tail: 0.007390564
8. t Critical one-tail: 1.649965767
9. [tex]\( P ( T \leq t ) \)[/tex] two-tail: 0.014781127
10. t Critical two-tail: 1.967929669
### Problem Parts:
1. The population parameter:
The population parameter is what we are testing against, which is the hypothesized mean, 3 hours per week.
[tex]\[
\text{Population Parameter} = \mu_0 = 3
\][/tex]
2. The sample statistic:
The sample statistic is the summary statistic from our sample data, which is the sample mean.
[tex]\[
\text{Sample Statistic} = \bar{x} = 3.158191551
\][/tex]
3. Randomness:
To determine if the data meets the randomness criterion, we need to know whether the sample was collected randomly. Usually, if the sample is a random sample, it is stated explicitly in the problem. Here it is stated that the sample was random.
[tex]\[
\text{Randomness criterion met} = \text{Yes}
\][/tex]
4. Normality:
With a large sample size (n >= 30), the Central Limit Theory states that the sampling distribution of the sample mean is approximately normal regardless of the shape of the population distribution.
[tex]\[
\text{Normality criterion met} = \text{Yes}
\][/tex]
5. The mean homework time for all students at the university:
This is the population parameter we are testing, which is the hypothesized mean.
[tex]\[
\text{Mean homework time for all students at the university} = \mu_0 = 3
\][/tex]
6. The proportion of all students that spend different than 3 hours on homework:
This refers to the two-tail p-value from the t-test, as it gives the probability of observing a sample statistic as extreme as the one observed, considering both directions from the hypothesized mean.
[tex]\[
\text{Proportion of all students that spend different than 3 hours on homework} = \text{Two-tail p-value} = 0.014781127
\][/tex]
### Conclusion
- The population parameter: [tex]\( 3 \)[/tex]
- The sample statistic: [tex]\( 3.158191551 \)[/tex]
- Randomness: Meets the criterion (Yes)
- Normality: Meets the criterion (Yes)
- The mean homework time for all students at the university: [tex]\( 3 \)[/tex]
- The proportion of all students that spend different than 3 hours on homework: [tex]\( 0.014781127 \)[/tex]
This solution follows from the given data and statistical principles without using or referencing any specific Python code execution.
