Answer:
1. To determine whether there is a significant difference between the population mean and the sample mean, we can use a one-sample t-test. The null hypothesis (H0) is that there is no significant difference between the population mean and sample mean, while the alternative hypothesis (Ha) is that there is a significant difference.
Using a significance level of 5%, and assuming the population standard deviation is unknown, we can calculate the t-statistic as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (24000 - μ) / (5000 / sqrt(25))
t = (24000 - μ) / 1000
At a significance level of 5% with 24 degrees of freedom (25-1), the critical t-value is ±2.064.
If the calculated t-value falls outside this range, then we reject the null hypothesis in favor of the alternative hypothesis.
Since the question does not provide a specific value for μ (population mean), we cannot complete the calculation to determine whether there is a significant difference between the population and sample mean.
2. To determine whether there is enough evidence that the average amount of active ingredient in the drug is greater than 12 mg, we can use a one-sample t-test. The null hypothesis (H0) is that the average amount of active ingredient is equal to 12 mg, while the alternative hypothesis (Ha) is that it is greater than 12 mg.
Using a confidence level of 99%, and assuming the population standard deviation is unknown, we can calculate the t-statistic as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (12.1 - 12) / (0.5 / sqrt(30))
t = 2.12
At a confidence level of 99% with 29 degrees of freedom (30-1), the critical t-value is 2.462.
Since the calculated t-value (2.12) is less than the critical t-value (2.462), we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the average amount of active ingredient in the drug is greater than 12 mg at a confidence level of 99%.
3. To test whether the transaction fee is higher than the remittance center's claim, we can use a one-sample t-test. The null hypothesis (H0) is that the mean transaction cost is equal to or lower than 3%, while the alternative hypothesis (Ha) is that it is higher than 3%.
Using a significance level of 10%, and assuming the population standard deviation is unknown, we can calculate the t-statistic as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (3.4 - 3) / (0.5 / sqrt(20))
t = 4.38
At a significance level of 10% with 19 degrees of freedom (20-1), the critical t-value is 1.734.
Since the calculated t-value (4.38) is greater than the critical t-value (1.734), we reject the null hypothesis and conclude that there is enough evidence to suggest that the transaction fee is higher than the remittance center's claim at a significance level of 10%.
4. To test whether the average hours spent on school work is less than 7 hours, we can use a one-sample t-test. The null hypothesis (H0) is that the mean number of hours spent on school work is equal to or greater than 7, while the alternative hypothesis (Ha) is that it is less than 7.
Using a confidence level of 95%, and assuming the population standard deviation is unknown, we can calculate the t-statistic as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (6.5 - 7) / (0.8 / sqrt(100))
t = -3.125
At a confidence level of 95% with 99 degrees of freedom (100-1), the critical t-value is -1.984.
Since the calculated t-value (-3.125) falls outside the range of the critical t-value (-1.984 to 1.984), we reject the null hypothesis and conclude that there is enough evidence to suggest that the average hours spent on school work is less than 7 hours at a confidence level of 95%.