So, what our professor does for tests is he gives us a list of questions to work through from which he will pull our exam questions. He recently posted the answer key to these questions, and one of them...I follow the first half of his solution, but his second half looks like it is riddled with errors (specifically the part where he is taking about the power of the test with the updated p=0.6).

The Superbowl ceremonial coin is suspected to be biased in favor of "heads". A statistician has decided to ip the coin n = 50 times in testing: H0: p = 0.5 versus H1: [0 > 0.5 where p represents the proportion of \"heads\" observed. a. What rejection rule should be used if we wish to control the probability of a Type I error at a = 0.05? b. Using the rejection rule for part a, what would be the power ofthis test if the coin was biased and the proportion of heads was actually 39 = 0.6? Answer a. We should check if n is sufficiently large before we choose our method. up i [3 x 11190 p)] = 50(. 5) i [50 X .5 x (1 .5)] = (14435.6). Since these values are between 0 and 50, we can use the normal approximation. As this is a right tailed test, with a = 0.05, we find the critical value to be qnorm[.95,0,1)= 1.6448536 . We can use this to solve for the value of X [number of heads) that would result in rejection using Z : X_np and p0 is the value of p assuming H0 is true. Solving for X, \\IPoUpo) we findX = npo + 1.644 X 1fnpo p0) = 50(. 5) + 1.644 x 1l(50(. 5)(1 .5) = 30.81 The power is 1-P[Type 11 Error) or P[Reject H0 | p : .6). Here, we reject H0 if the number of heads (X) is greater than 30.81 when p = .6 or the number of heads is up = 50(. 6) = 30. When p = .6, the standard error is 1/1np(1 p = 50(. 6)(. 4) = 4.47. This means the power is $P(X>30.81| ,)=$0.4197629