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Sagot :
Let's analyze this problem step by step to determine if the data meets certain statistical conditions and if the university's claim can be tested using the given sample data.
### Problem Statement and Hypotheses
The university claims that students spend a mean of 3 hours per week on homework for every credit hour of class. The administration surveyed 300 students to check if this mean has changed, with a significance level of 0.05.
### Given Data
- Mean of sample ([tex]\(\overline{x}\)[/tex]): 3.158191551
- Hypothesized Mean ([tex]\(\mu\)[/tex]): 3
- Variance of sample ([tex]\(s^2\)[/tex]): 1.248743026
- Sample size ([tex]\(n\)[/tex]): 300
- Degrees of freedom ([tex]\(df\)[/tex]): 299
- [tex]\(t\)[/tex]-statistic: 2.451926089
- [tex]\(P(T \le t)\)[/tex] one-tail: 0.007390564
- [tex]\(P(T \le t)\)[/tex] two-tail: 0.014781127
- [tex]\(t\)[/tex] Critical one-tail: 1.649965767
- [tex]\(t\)[/tex] Critical two-tail: 1.967929669
### Population Parameter and Sample Statistic
- The population parameter is the hypothesized mean homework time, which is [tex]\(\mu = 3\)[/tex] hours.
- The sample statistic is the mean homework time from the sample, which is [tex]\(\overline{x} = 3.158191551\)[/tex] hours.
### Conditions for [tex]\(t\)[/tex]-Distribution
1. Randomness: The data should be collected randomly.
- Given Information: Randomness condition is met (True).
2. Normality: For large samples (n > 30), the Central Limit Theorem allows us to assume the sample mean is normally distributed.
- Given Information: The sample size [tex]\(n = 300\)[/tex] is greater than 30, so Normality condition is met (True).
### Hypothesis Testing
We will perform a two-tailed test since the claim is about the mean being different from 3 hours, not specifically greater or less than 3 hours.
#### Null and Alternative Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 3\)[/tex] hours
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu \neq 3\)[/tex] hours
#### Test Statistic:
The test statistic given is [tex]\(t = 2.451926089\)[/tex].
#### Decision Rule:
- For a significance level of [tex]\(\alpha = 0.05\)[/tex]:
- The critical t-value for a two-tailed test with 299 degrees of freedom is approximately [tex]\( \pm 1.967929669 \)[/tex].
#### P-Values:
- One-tailed [tex]\(P\)[/tex]-value: 0.007390564
- Two-tailed [tex]\(P\)[/tex]-value: 0.014781127
#### Conclusion:
- Since the [tex]\(t\)[/tex]-statistic (2.451926089) is greater than the critical t-value for the two-tailed test (1.967929669) and since the two-tailed p-value (0.014781127) is less than our significance level (0.05), we reject the null hypothesis.
- Therefore, there is sufficient evidence to conclude that the mean homework time is significantly different from 3 hours at the 0.05 significance level.
### Summary
- Population parameter: Hypothesized Mean = 3 hours.
- Sample statistic: Mean of the sample = 3.158191551 hours.
- Randomness condition met: True.
- Normality condition met: True.
Thus, based on the given data and statistical analysis, we conclude that the mean homework time for students is significantly different from the hypothesized 3 hours per week.
### Problem Statement and Hypotheses
The university claims that students spend a mean of 3 hours per week on homework for every credit hour of class. The administration surveyed 300 students to check if this mean has changed, with a significance level of 0.05.
### Given Data
- Mean of sample ([tex]\(\overline{x}\)[/tex]): 3.158191551
- Hypothesized Mean ([tex]\(\mu\)[/tex]): 3
- Variance of sample ([tex]\(s^2\)[/tex]): 1.248743026
- Sample size ([tex]\(n\)[/tex]): 300
- Degrees of freedom ([tex]\(df\)[/tex]): 299
- [tex]\(t\)[/tex]-statistic: 2.451926089
- [tex]\(P(T \le t)\)[/tex] one-tail: 0.007390564
- [tex]\(P(T \le t)\)[/tex] two-tail: 0.014781127
- [tex]\(t\)[/tex] Critical one-tail: 1.649965767
- [tex]\(t\)[/tex] Critical two-tail: 1.967929669
### Population Parameter and Sample Statistic
- The population parameter is the hypothesized mean homework time, which is [tex]\(\mu = 3\)[/tex] hours.
- The sample statistic is the mean homework time from the sample, which is [tex]\(\overline{x} = 3.158191551\)[/tex] hours.
### Conditions for [tex]\(t\)[/tex]-Distribution
1. Randomness: The data should be collected randomly.
- Given Information: Randomness condition is met (True).
2. Normality: For large samples (n > 30), the Central Limit Theorem allows us to assume the sample mean is normally distributed.
- Given Information: The sample size [tex]\(n = 300\)[/tex] is greater than 30, so Normality condition is met (True).
### Hypothesis Testing
We will perform a two-tailed test since the claim is about the mean being different from 3 hours, not specifically greater or less than 3 hours.
#### Null and Alternative Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 3\)[/tex] hours
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu \neq 3\)[/tex] hours
#### Test Statistic:
The test statistic given is [tex]\(t = 2.451926089\)[/tex].
#### Decision Rule:
- For a significance level of [tex]\(\alpha = 0.05\)[/tex]:
- The critical t-value for a two-tailed test with 299 degrees of freedom is approximately [tex]\( \pm 1.967929669 \)[/tex].
#### P-Values:
- One-tailed [tex]\(P\)[/tex]-value: 0.007390564
- Two-tailed [tex]\(P\)[/tex]-value: 0.014781127
#### Conclusion:
- Since the [tex]\(t\)[/tex]-statistic (2.451926089) is greater than the critical t-value for the two-tailed test (1.967929669) and since the two-tailed p-value (0.014781127) is less than our significance level (0.05), we reject the null hypothesis.
- Therefore, there is sufficient evidence to conclude that the mean homework time is significantly different from 3 hours at the 0.05 significance level.
### Summary
- Population parameter: Hypothesized Mean = 3 hours.
- Sample statistic: Mean of the sample = 3.158191551 hours.
- Randomness condition met: True.
- Normality condition met: True.
Thus, based on the given data and statistical analysis, we conclude that the mean homework time for students is significantly different from the hypothesized 3 hours per week.
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