# Statistics Hypothesis Testing + binomial table problem

There is a problem in my statistics textbook (Wackerly 7th) I cannot seem to solve. The question is:

10.8 A two-stage clinical trial is planned for testing $H_0 : p = .10$ versus $H_a : p > .10$, where $p$ is the proportion of responders among patients who were treated by the protocol treatment. At the first stage, 15 patients are accrued and treated. If 4 or more responders are observed among the (first) 15 patients, $H_0$ is rejected, the study is terminated, and no more patients are accrued. Otherwise, another 15 patients will be accrued and treated in the second stage. If a total of 6 or more responders are observed among the 30 patients accrued in the two stages (15 in the first stage and 15 more in the second stage), then H0 is rejected. For example, if 5 responders are found among the first-stage patients, $H_0$ is rejected and the study is over. However, if 2 responders are found among the first-stage patients, 15 second-stage patients are accrued, and an additional 4 or more responders (for a total of 6 or more among the 30) are identified, $H_0$ is rejected and the study is over.

a) Use the binomial table to find the numerical value of $α$ for this testing procedure.

b) Use the binomial table to find the probability of rejecting the null hypothesis when usingvthis rejection region if $p = .30$.

c) For the rejection region defined above, find $β$ if $p = .30$.

My attempt to solve part a is as follows:

Let $X$ be the number of responders from the 1st stage, and $Y$ be the number of responders in the 2nd stage. Then,

$n = 15, p=0.10$

$α = P(X \geq 4) + \sum_{i=0}^3P(Y\geq6-i)P(X\leq i)$

$α = 0.056+(0.002)(0.206)+(0.013)(0.549)+(0.056)(0.816)+(0.184)(0.944)$

$α = 0.283$

The answer in the back of the book says $α=0.0989$, so either I'm reading the table wrong or the book is wrong. I would appreciate any input.

Thanks.

Your calculation for part (a) is not correct because the formula should read $$\alpha = \Pr[X \ge 4] + \sum_{i=0}^3 \Pr[Y \ge 6-i]\Pr[X = i],$$ not $\Pr[X \le i]$. Therefore, the table of probabilities will be $$\begin{array}{ll} \Pr[X = 0] \approx 0.205891 & \Pr[Y \ge 6] \approx 0.00224967 \\ \Pr[X = 1] \approx 0.343152 & \Pr[Y \ge 5] \approx 0.0127205 \\ \Pr[X = 2] \approx 0.266896 & \Pr[Y \ge 4] \approx 0.0555556 \\ \Pr[X = 3] \approx 0.128505 & \Pr[Y \ge 3] \approx 0.184061 \\ \Pr[X \ge 4] \approx 0.0555556. & \end{array}$$