Hypothesis Testing
I believe that dogs are as smart as people. Assume IQ of a dog follows [math]X_i \sim N(\mu,10^2)[/math]. IQ of 10 dogs are measured: 30, 25, 70, 110, 40, 80, 50, 60, 100, 60. We want to test if dogs are as smart as people by testing
One reasonable thing one may try is to see how high the sample mean is.
> x<-c(30, 25, 70, 110, 40, 80, 50, 60, 100, 60)
> mean(x)
[1] 62.5
Since the average IQ of 10 dogs are lower than 100, one would be inclined to reject [math]H_0[/math].
Let [math]\bar{X} [/math] be a test statistic and [math]R = (−∞,90][/math] to be a rejection region. Let’s compute the probability of making Type I error based on this testing procedure. Under the assumption [math]H_0[/math] is true,
Under this condition, [math]\bar{X} \sim N(100, 10)[/math] and
> pnorm(90,100,sqrt(10))
[1] 0.0007827011
By using this test procedure, it is highly unlikely to make Type I error. Let’s see what happens when we change the rejection region.
When [math]R = (−∞,95], \alpha = P(\bar{X} \leq 95) [/math].
> pnorm(95,100,sqrt(10))
[1] 0.05692315
When [math]R = (−∞,99], \alpha = P(\bar{X} \leq 99) [/math].
> pnorm(99,100,sqrt(10))
[1] 0.3759148
The test procedure based on rejecting [math]H_0 \text{ if } \bar{X} \leq 99[/math] will produce huge Type I error.
Decision | H0 false | H0 true |
---|---|---|
Reject H0 | Correct. We can be 95% sure that we made a right decision, because in our case; ([math]p = 1 - \alpha [/math]) 1 – .05 = .95. |
Type I error = a level, p = a, probability of error is commonly set at .05. |
Fail to reject H0 | Type II error, b level = maximum accepted probability is suggested to be set to .20. | Correct. Probability: p = 1 – b. In our case 1 – .20 = .80. We would make a right decision based on our analyses 80% of the time. |
References
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