Main Purpose
Based on sample data mean ~X, examine the the population mean \(\mu\)
Scenario
The standard of one beverage can is 500, and is a normal distribution, X ~ N (500,102) Now someone doubts that this average size has down to 490, and he did sample test for 9 subjects, during which sample Expected vaalue is 492. Based on this, we will examine whether this is true or not to say the size has down.
Examine these 2 hypothesis: \[H_{0}: \mu = 500\]
\[H_{1}: \mu = 490\]
Discussion
For test statistic (sample mean ~X) smaller, it is great for H1, therefore, if H1 is true, i.e. H0 is false,we reject the H0. Then ~X should be small enough. Thus, we should have a standard to reflect this property.
Significance Standard and Rejection Region
Assume ~X is smaller than one constant, C, then we say that it falls in the rejection region, we reject H0 \[P(～X < C) \leq \alpha\] We say that is a standard which is usually be 0.05, and the whole meaning is: Under the allowed maximal rate of error, H0 is rejected. Type Errors
Because we cannot completely say that our conjuction is absolutely true, so there is probability of error, here is 2 kinds errors
1.Type I: Reject the truth.
If H0 is the truth, and our decision is H1, that is equal to we reject the truth.
2.Type II: Accept the error.
If H0 is false, i.e. H1 is the truth, and our decision is H0, that is equal to we accept the error
Solution
The main purpose of this test is to compare sample mean and C, that is to say, the core step is calculation of C. Based on 5%, and ~X is normally distributed, so we can come out C by qnorm( ) in R.
qnorm(0.05)## [1] -1.644854So theoretical C should be 1.645 away from 0. The adopted test statistic is like this: \[ T = \frac{～X - \mu}{s/\sqrt{n}}\]
3*(492-500)/10## [1] -2.4The test statistic observed is -2.4, while critical value -1.645, the former is smaller. Thus, H0 is false, and H1 is true.
Conslusion
The conjuction is right, the can size has down to 90.
H0 is rejected, H1 is accepted.