1. Mean of variable 1 = 429.4 g
This is the average values for the control group
2. Mean of variable 2 = 413.6 g
This is the average value of the tested group.
3. P(T <= t) one-tail
The number 0.12639 (or 12.6%) is the p-value of the mean 413.6 if we are to assume that the real average is the control group average of 429.4 g.
Note:
We have chosen the t-test with equal variance for this scenario. The reason for choosing equal variance is because both set of data comes from the same batch of rat. The standard deviation of the rat should be the same be it the control group or the tested group.
shown.
If lemon juice has neutral effect on weight loss, all the averages calculated from a sample of 10 rats should be distributed as shown with 429.4 at the centre. The average that is calculated from our tested group is 413.6 g which is at the position shown.
From our initial setting hypothesis, we have the following:
Null Hypothesis: Lemon has no significant effect on weight loss
Alternative Hypothesis: Lemon has significant effect on weight loss
If we are to reject the Null Hypothesis based on the averages from our control and tested group, we will have a 12.6% chance of committing Type 1 error
[In another word, if we based on the averages from our control and tested group and claim that lemon is effective in weight loss, we will have a 12.6% change of saying the wrong thing and make ourselves look like fool.]
429.4 413.6
1. Mean of variable 1 = 4 mg
This is the claimed value by the pharmaceutical company.
2. Mean of variable 2 = 3.99235 mg
This is the average value of the randomly selected sample.
3. P(T <= t) one-tail
The number 0.00027 (or 0.027% ) is the p-value of the mean 3.99235 mg if we are to assume that the real average is the claimed value of 4 mg.
Note:
We have chosen the t-test with unequal variance for this scenario. The data in variable 1 are all the same value (i.e. 4 mg) while the data in variable 2 are the collected data. The word variance is almost equivalent to standard deviation. Since the data in variable 1 is fixed while the data in variable 2 vary, the standard deviation (or variance) has to be unequal
shown.
If the active ingredient in the tablet is 4 mg as claimed, all the averages calculated from a sample of 20 tablets be distributed as shown with 4 at the centre. The average that is calculated from our collected sample is 3.99235 mg which is at the position shown.
From our initial setting hypothesis, we have the following:
Null Hypothesis: The pharmaceutical company is honest with the claim (ie 4 mg)
Alternative Hypothesis: The pharmaceutical company is dishonest with the claim (ie not 4 mg)
If we are to reject the Null Hypothesis based the average obtained from our sample, we will have a 0.027% chance of committing Type 1 error
[In another word, if we based on the averages obtained from our sample and claim that the pharmaceutical company is making a false claim, we will have a 0.027% chance of wrongfully accusing the company of lying.]
4 3.99235
1. Mean of variable 1 = 200 g
This is the claimed supporting strength of the thread by the fashion company.
2. Mean of variable 2 = 193.533 g
This is the average strength of the thread from our randomly selected sample.
3. P(T <= t) one-tail
The number 0.08086 (or 8.09%) is the p-value of the mean 193.533 mg if we are to assume that the real average is the claimed value of 200 g.
Note:
We have chosen the t-test with unequal variance for this scenario. The data in variable 1 are all the same value (i.e. 200 g) while the data in variable 2 are the collected data. The word variance is almost equivalent to standard deviation. Since the data in variable 1 is fixed while the data in variable 2 vary, the standard deviation (or variance) has to be unequal
shown.
If the strength of the thread is 200 g as claimed, all the averages calculated from a sample of 6 threads will be distributed as shown with 200 at the centre. The average that is calculated from our collected sample is 193.53 g which is at the position shown.
From our initial setting hypothesis, we have the following:
Null Hypothesis: The fashion company is honest with the claim (ie the strength of the thread can withstand at least 200 g)
Alternative Hypothesis: The fashion company is dishonest with the claim (ie the strength of the thread cannot withstand 200 g)
If we are to reject the Null Hypothesis based the average obtained from our sample, we will have a 8.09% chance of committing Type 1 error
[In another word, if we based on the averages obtained from our sample and claim that the fashion company is making a false claim, we will have a 8.09% chance of wrongfully accusing the company of lying.]
200 193.53