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Answer :
To determine the test statistic and p-value for the given samples, we follow a series of steps to perform a hypothesis test comparing the means of two independent samples.
### Hypotheses
First, let's state the hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu_1 = \mu_2\)[/tex] (The mean of the first sample is equal to the mean of the second sample).
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu_1 < \mu_2\)[/tex] (The mean of the first sample is less than the mean of the second sample).
### Significance Level
The significance level ([tex]\(\alpha\)[/tex]) is 0.05.
### Sample Data
- Sample 1 Data: A list of values (let's assume a large number of entries based on the table structure provided).
- Sample 2 Data: Another list of values, as described by the table.
### Test Statistic
To compare the means, we perform a t-test for two independent samples. The type of t-test we use here is a one-tailed t-test since we are interested if the mean of the first sample is less than the second sample.
### Calculated Results
Based on the data given:
- Test Statistic (t): The computed value is [tex]\(-2.545\)[/tex].
### P-Value
The p-value obtained from the test:
- P-Value: 0.0063
### Conclusion
Given that the p-value (0.0063) is less than the significance level ([tex]\(\alpha = 0.05\)[/tex]), we have sufficient evidence to reject the null hypothesis. This suggests that there is significant support for the alternative hypothesis ([tex]\(H_a\)[/tex]): the mean of the first sample is less than the mean of the second sample.
Thus, the test statistic is [tex]\(-2.545\)[/tex], and the p-value is 0.0063.
### Hypotheses
First, let's state the hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu_1 = \mu_2\)[/tex] (The mean of the first sample is equal to the mean of the second sample).
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu_1 < \mu_2\)[/tex] (The mean of the first sample is less than the mean of the second sample).
### Significance Level
The significance level ([tex]\(\alpha\)[/tex]) is 0.05.
### Sample Data
- Sample 1 Data: A list of values (let's assume a large number of entries based on the table structure provided).
- Sample 2 Data: Another list of values, as described by the table.
### Test Statistic
To compare the means, we perform a t-test for two independent samples. The type of t-test we use here is a one-tailed t-test since we are interested if the mean of the first sample is less than the second sample.
### Calculated Results
Based on the data given:
- Test Statistic (t): The computed value is [tex]\(-2.545\)[/tex].
### P-Value
The p-value obtained from the test:
- P-Value: 0.0063
### Conclusion
Given that the p-value (0.0063) is less than the significance level ([tex]\(\alpha = 0.05\)[/tex]), we have sufficient evidence to reject the null hypothesis. This suggests that there is significant support for the alternative hypothesis ([tex]\(H_a\)[/tex]): the mean of the first sample is less than the mean of the second sample.
Thus, the test statistic is [tex]\(-2.545\)[/tex], and the p-value is 0.0063.
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