To answer these questions, we can use the standard normal distribution table or a statistical calculator. The calculations involve finding critical values based on the given significance level (alpha) and the z-score associated with the given Type I or Type II error.
1. Cut-off point at alpha = 0.05:
Using the standard normal distribution table or a calculator, the z-score corresponding to an alpha of 0.05 (two-tailed test) is approximately 1.96. Therefore, the cut-off point is 1.96 standard deviations away from the mean.
2. Cut-off point if Type I error is 1%:
For a Type I error of 1% (alpha = 0.01, two-tailed test), the corresponding z-score can be found using the standard normal distribution table or a calculator. The z-score is approximately 2.58.
3. Rejecting the Null at 5% level of significance:
To determine whether to reject the null hypothesis at the 5% level of significance, we compare the sample mean delivery time to the expected delivery time stated by the company. If the sample mean falls within the range of values that are unlikely to occur by chance (i.e., beyond the cut-off point), we would reject the null hypothesis. In this case, the sample mean is 16.6 days, which is within the cut-off range of 1.96 standard deviations. Therefore, we would not reject the null hypothesis at the 5% level of significance.
4. Rejecting the Null at 1% level of significance:
Similar to the previous question, we compare the sample mean delivery time to the expected delivery time stated by the company. If the sample mean falls within the range of values that are unlikely to occur by chance (beyond the cut-off point), we would reject the null hypothesis. In this case, the sample mean is still 16.6 days, which is within the cut-off range of 2.58 standard deviations. Therefore, we would not reject the null hypothesis at the 1% level of significance.
5. Type II error at Alpha = 0.01 (assuming true delivery time is 17 days):
To calculate the Type II error, we need to determine the area under the standard normal distribution curve that corresponds to failing to reject the null hypothesis when it is false. This is the probability of accepting the null hypothesis (not detecting a difference) when there is a true difference in the population. Given that the true delivery time is 17 days, we need to find the probability that the sample mean falls within the cut-off range. Using the known population standard deviation of 5.6 days, we can calculate the z-score:
z = (sample mean - true mean) / (standard deviation / sqrt(sample size))
z = (16.6 - 17) / (5.6 / sqrt(49))
z = -0.4 / (5.6 / 7)
z = -0.4 / 0.8
z = -0.5
Using the standard normal distribution table or a calculator, the probability associated with a z-score of -0.5 is approximately 0.3085. Therefore, the Type II error at alpha = 0.01 is 0.3085.
6. Type II error at Alpha = 0.05 (assuming true delivery time is 17 days):
Using the same calculation method as in question 5, we find the z-score for a sample mean of 16.6 days when the true mean is 17 days. The z-score is -0.5. Using the standard normal distribution table or a calculator, the probability associated with a z-score of -0.5 is approximately 0.308