Answer :
### Step-by-Step Solution
In this problem, we are given data from a sample of 300 students and are testing the hypothesis that the mean number of hours spent on homework is different from the claimed mean of 3 hours per week per credit hour.
Let's outline the solution step by step:
Step 1: State the hypotheses
We are conducting a two-tailed test with the following hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 3\)[/tex]
- The mean homework time is 3 hours per week per credit hour.
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu \neq 3\)[/tex]
- The mean homework time is different from 3 hours per week per credit hour.
Step 2: Gather the sample statistics
From the problem, we have:
- Sample size ([tex]\(n\)[/tex]) = 300
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 3.158191551
- Sample variance (s²) = 1.248743026
- Hypothesized mean ([tex]\(\mu_0\)[/tex]) = 3
- Degrees of freedom ([tex]\(df\)[/tex]) = [tex]\(n - 1 = 299\)[/tex]
Step 3: Identify the test statistic and its value
The test statistic used is the t-statistic for a one-sample t-test. The given t-statistic is:
- [tex]\( t \text{ stat} = 2.451926089 \)[/tex]
Step 4: Determine the significance level and the critical t-value
We are given the significance level ([tex]\(\alpha\)[/tex]) = 0.05.
For a two-tailed test:
- The critical t-value for a 0.05 significance level with [tex]\(df = 299\)[/tex] is:
- [tex]\(t \text{ critical two-tail} = \pm1.967929669\)[/tex]
Step 5: Calculate the p-value for the test statistic
- The provided p-value for the two-tailed test is:
- [tex]\(p \text{ value two-tail} = 0.014781127\)[/tex]
Step 6: Decision Rule
We compare the calculated t-statistic with the critical t-value, and the p-value with the significance level:
- If [tex]\( |t| > t \text{ critical two-tail} \)[/tex] or if [tex]\( p \value \leq \alpha\)[/tex], we reject the null hypothesis.
Step 7: Conclusion
- Since [tex]\(t = 2.451926089\)[/tex] > [tex]\(1.967929669\)[/tex] and [tex]\(p \text{ value two-tail} = 0.014781127 < 0.05\)[/tex], we reject the null hypothesis.
### Verifying Conditions for T-Distribution
Randomness Condition:
- It is stated that the sample was randomly selected, so the randomness condition is met.
Normality Condition:
- The Central Limit Theorem states that for sufficiently large sample sizes (typically [tex]\(n \geq 30\)[/tex]), the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution. Since [tex]\(n = 300\)[/tex], which is significantly large, the normality condition is satisfied.
### Final Answer
Given the t-statistic and p-value, we have sufficient evidence to reject the null hypothesis at the 0.05 significance level. This means that the mean homework time is statistically significantly different from 3 hours per week per credit hour.
Does the data meet the conditions for the t-distribution?
- Randomness: Yes, as stated.
- Normality: Yes, because the sample size is 300, which is greater than 30, making the sampling distribution of the sample mean approximately normal.
Therefore, the final conclusion is:
Check Yes, because the population was stated to be normal.
This thorough step-by-step solution ensures that the correct conclusion is reached based on the provided statistics and hypothesis testing principles.
In this problem, we are given data from a sample of 300 students and are testing the hypothesis that the mean number of hours spent on homework is different from the claimed mean of 3 hours per week per credit hour.
Let's outline the solution step by step:
Step 1: State the hypotheses
We are conducting a two-tailed test with the following hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 3\)[/tex]
- The mean homework time is 3 hours per week per credit hour.
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu \neq 3\)[/tex]
- The mean homework time is different from 3 hours per week per credit hour.
Step 2: Gather the sample statistics
From the problem, we have:
- Sample size ([tex]\(n\)[/tex]) = 300
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 3.158191551
- Sample variance (s²) = 1.248743026
- Hypothesized mean ([tex]\(\mu_0\)[/tex]) = 3
- Degrees of freedom ([tex]\(df\)[/tex]) = [tex]\(n - 1 = 299\)[/tex]
Step 3: Identify the test statistic and its value
The test statistic used is the t-statistic for a one-sample t-test. The given t-statistic is:
- [tex]\( t \text{ stat} = 2.451926089 \)[/tex]
Step 4: Determine the significance level and the critical t-value
We are given the significance level ([tex]\(\alpha\)[/tex]) = 0.05.
For a two-tailed test:
- The critical t-value for a 0.05 significance level with [tex]\(df = 299\)[/tex] is:
- [tex]\(t \text{ critical two-tail} = \pm1.967929669\)[/tex]
Step 5: Calculate the p-value for the test statistic
- The provided p-value for the two-tailed test is:
- [tex]\(p \text{ value two-tail} = 0.014781127\)[/tex]
Step 6: Decision Rule
We compare the calculated t-statistic with the critical t-value, and the p-value with the significance level:
- If [tex]\( |t| > t \text{ critical two-tail} \)[/tex] or if [tex]\( p \value \leq \alpha\)[/tex], we reject the null hypothesis.
Step 7: Conclusion
- Since [tex]\(t = 2.451926089\)[/tex] > [tex]\(1.967929669\)[/tex] and [tex]\(p \text{ value two-tail} = 0.014781127 < 0.05\)[/tex], we reject the null hypothesis.
### Verifying Conditions for T-Distribution
Randomness Condition:
- It is stated that the sample was randomly selected, so the randomness condition is met.
Normality Condition:
- The Central Limit Theorem states that for sufficiently large sample sizes (typically [tex]\(n \geq 30\)[/tex]), the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution. Since [tex]\(n = 300\)[/tex], which is significantly large, the normality condition is satisfied.
### Final Answer
Given the t-statistic and p-value, we have sufficient evidence to reject the null hypothesis at the 0.05 significance level. This means that the mean homework time is statistically significantly different from 3 hours per week per credit hour.
Does the data meet the conditions for the t-distribution?
- Randomness: Yes, as stated.
- Normality: Yes, because the sample size is 300, which is greater than 30, making the sampling distribution of the sample mean approximately normal.
Therefore, the final conclusion is:
Check Yes, because the population was stated to be normal.
This thorough step-by-step solution ensures that the correct conclusion is reached based on the provided statistics and hypothesis testing principles.
