10.6 The proportion of adults living in a small town who are college graduates is estimated to be p = 0.6. To test this hypothesis, a random sample of 15 adults is selected. If the number of college graduates in the sample is anywhere from 6 to 12, we shall not reject the null hypothesis that p = 0.6; otherwise, we shall conclude that p = 0.6. = 0.6. Use the binomial distribution. (a) Evaluate a assuming that p (b) Evaluate for the alternatives p = (c) Is this a good test procedure? 0.5 and p = 0.7. 10.7 Repeat Exercise 10.6 but assume that 200 adults are selected and the fail-to-reject region is defined to be 110 ≤ x ≤ 130, where is the number of college graduates in our sample. Use the normal approximation.

It is given that: The sample size, n=200. The claimed proportion of college graduates among the adults living in the area, i.e., p0=0.6. Part (a) Considering np0=200×0.6=120 ≥5 and n1-p0=200×1-0.6=80 ≥5, the normal approximation for the distribution of the proportion of college graduates among adults is inferred to be possible. For 200 sampled adults, the mean of the sampling distribution of the proportion of college graduates, i.e., μp^=p0=0.6. And the standard deviation of the sampling distribution of the proportion of college graduates, i.e., σp^=p0 1-p0n=0.6×1-0.4200≈0.03464. For the defined fail-to-reject region, the likelihood of not rejecting the hypothesis that p=0.6, the respective likelihood is: P110≤x≤130=P110200≤xn≤130200=P0.55≤p^≤0.65=P0.55-0.60.03464≤p^-μ0σp^≤0.65-0.60.03464≈P-1.4434≤Z≤1.4434≈PZ≤1.4434-PZ<-1.4434 Using the excel function, the area to the left of the z-scores of 1.4434 and -1.4434 are obtained as follows: So using PZ≤1.4434≈0.9255 and PZ<-1.4434≈0.0745, the area (in terms of probability) of the fail-to-reject region is: Pfail-to-reject=P110≤x≤130≈PZ≤1.4434-PZ<-1.4434≈0.9255-0.0745≈0.851 So the likelihood of committing a type-I error, i.e., α=1-Pfail-to-reject≈1-0.851≈0.149 So the evaluated value of α is 0.149 or 14.9%. Part (b) For z-critical values for which H0 is not rejected are -1.4434 and 1.4434. So H0 is not rejected if the value of p^ (i.e., the sample proportion of college graduate adults) lies within p0+-zα2σp^ and p0+zα2σp^. So the H0 is not rejected if, p0+-zα2σp^≤p^≤p0+-zα2σp^=0.6+-1.4434×0.03464≤p^≤0.6+1.4434×0.03464≈0.55≤p^≤0.65 For p=0.5, the probability of type-II error, i.e., β=Ptype-II error=PDo not reject H0|p=0.5=P0.55≤p^≤0.65|p=0.5=P0.55-0.50.5×1-0.5200≤p^-pp×1-pn≤0.65-0.50.5×1-0.5200≈P1.4142≤Z≤4.2426≈PZ≤4.2426-PZ<1.4142 Using the excel function the area to the left of the z-scores of 4.2426 and 1.4142 are obtained as follows: Using PZ≤4.2426≈1.0000 and PZ<1.4142≈0.9213, it is obtained that: β=Ptype-II error≈PZ≤4.2426-PZ<1.4142≈1.0000-0.9213≈0.0787 And for p=0.7, the probability of type-II error, i.e., β=Ptype-II error=PDo not reject H0|p=0.7=P0.55≤p^≤0.65|p=0.7=P0.55-0.70.7×1-0.7200≤p^-pp×1-pn≤0.65-0.70.7×1-0.7200≈P-4.629≤Z≤-1.543≈PZ≤-1.543-PZ<-4.629 Using the excel function the area to the left of the z-scores of -1.543 and -4.629 are obtained as follows: Using PZ≤-1.543≈0.0614 and PZ<-4.269≈0.0000, it is obtained that: β=Ptype-II error≈PZ≤-1.543-PZ<-4.629≈0.0614-0.0000≈0.0614 So the β values are 0.0787 and 0.0614 for the alternatives p=0.5 and p=0.7 respectively. Part (c) The power of the test for p=0.5 is 1-βp=0.5=1-0.0787=0.9213 i.e., 92.13%. And the power of the test for p=0.7 is 1-βp=0.7=1-0.0614=0.9386 i.e., 93.86%. Since a claim related to the proportion of college graduate adults is empirically testable with a statistically proven conclusion without any ambiguity the conclusion can be validated at a population level. Moreover, since the power of the test is comparatively higher, the likelihoods of not committing an error of type II are 92.13% and 93.86% for the provided true proportion values of 0.5 and 0.7 respectively. So this can be interpreted to be a good test procedure. Although the error of type-I probability is comparatively higher, i.e., 14.9%, a higher sample size may be considered in order to reduce this sampling error.