Question

Example 8.4 An automobile model is known to sustain no visible damage 25% of the time in 10-mph crash tests. A modified bumper design has been proposed in an effort to increase this percentage. Let p denote the proportion of all 10-mph crashes with this new bumper that result in no visible damage. The hypotheses to be tested are Ho: P = 0.25 (no improvement) versus Ha: p ? ? 0.25. The test will be based on an experiment involving n-20 independent crashes with prototypes of the new design. The natural test statistic here is X- the number of crashes with no visible damage. If Ho is true, E(X) = nPo = (20)('25) = 5. Intuition suggests that an observed value x much larger than this would provide strong evidence against Ho and in support of Ha Consider using a significance level of 0.10. The P-value is P(X 2 x when X has a binomial distribution with n = 20 and p 0.25) 1-6(x-1: 20, 0.25) for x > 0 Appendix Table A.1 shows that in this case, P(X 7) = 1-6(6; 20, 0.25) = 1-0.786 = 0.214 8) 1-0.898 = 0.102 0.10, P(X 9) = 1-0.959 = 0.041 P(X Thus rejecting Ho when P-value š 0.10 is equivalent to rejecting Ho when X 2 .Therefore P(committing a type I error) = P(rejecting Ho when Ho is true) - P(X 2 8 when X has a binomial distribution with n- and p = 0.25) (rounded to two decimal places). That is, the probability of a type 1 error is just the significance level ?. If the null hypothesis is true here and the test procedure is used over and over again, each time in conjunction with a group of 20 crashes, in the long run the null hypothesis will be incorrectly rejected in favor of the alternative hypothesis about 10% of the time, so our test procedure offers reasonably good protection against committing a type I erron There is only one type I error probability because there is only one value of the parameter for which Ho is true (this is one benefit of simplifying the null hypothesis to a claim of equality). Let B denote the probability of committing a type II error, unfortunately there is not a single value of ?, because there are a multitude of ways for Ho to be false- it could be false because p = 0.30, because p = 0.37, because p = 0.5, and so on. There is in fact a different value of ? for each different value of p that exceeds 0.25. At the chosen significance level 0.10, Ho ill be rejected if and only if X 2 8, so Ho will not be rejected if and only if X 7. Thus = P(type II error when p = 0.3) P(Ho is not rejected when p 0.3) = P(X 7 when X ~ Bin(20, 0.3)] B(7; 20, 0.3)0.772 when p is actually 0.3 rather than 0.25 (a "small" departure from Ho), roughly 77% of all experiments of this type would result in Ho being incorrectly not rejected! The accompanying table displays B for selected values of p (each calculated as we just did for B(0.3)). Clearly, B decreases as the value of p moves farther to the right of the null value 0.25. Intuitively, the greater the departure from Ho, the more likely it is that such a departure will be detected 0.3 0.4 0.5 0.6 0.7 0.8 B(P) 0.772 0.416 0.132 0.021 0.001 0.000 The probability of committing a type II error here is quite large when p 0.3 or 0.4. This is because those values are quite close to what Ho asserts and the sample size of 20 is too small to permit accurate discrimination between 0.25 and those values of p The proposed test procedure is still reasonable for testing the more realistic null hypothesis that p s 0.25. In this case, there is no longer a single type 1 error probability of ?, but instead there is an ? for each p that is at most 0.25: a(0.25), a(0.23), ?(0.20), a(0.15), and s It is easily verified, though, that ?(p) < ?(0.25) = 0.102 if p < 0.25. Tha the largest type I error probability occurs for the boundary value 0.25 between Ho and Ha. Thus if ? is small for the im ed WI ma ma for the m

Answer 1

(i)

$H_{a}:p>0.25$

(ii) Rejecting H₀ when p-value <=0.10 is equivalent to rejecting H₀ when,

$X \geq8$

(iii) P(committing a Type I error )= P(X >=8 when X has a binomial distribution with n=20 and p=0.25)

  P(committing a Type I error ) $\approx 0.04$

(iv)   $\beta (p=0.3)$