Question

2. Suppose we want to test whether a coin is fair (that is, the probability of heads is p = .5). We toss the coin 1000 times, and record the number of heads. Let T denote the number of heads divided by 1000. Consider a test that rejects the null hypothesis that p=.5 if T > c. (a) Write down a formula for P(T>c) assuming p = 0.5. (This formula may be compli- cated, but try to give an explicit expression.) (b) Give a large-sample approximate formula for PT > c) assuming p = 0.5, based on the central limit theorem. (c) Using the large-sample approximation, calculate the appropriate value c for a test at the 0.05 percent level. You can use the fact that if Z~N(0,1), then P(Z > 1.645) = 0.05. Would your test reject the null hypothesis if we observed 560 heads and 440 tails?

Answer 1

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where x is very small number usually known as significance level or size of the test. Detailed Exeplanation only for you dear student If you are satisfied please THUMBS UP and your valuable support is very important. Thanks... d = 0.1 OY 0.05 OY 0:01 Genesally for testing a simple null by pothesis, Vs Simple alterner tive hypothesis we can use Neymann - Pearson Most powerful test Now Back to our problem, het x denotes number of heads out of 1000 toss. a. x.. Binomial (1000, p) where p is probability of occurance of head. Null hypothesis Ho : p=0.5. - under Ho ; xoo Bin (1000, 05) het and T be be e the the test statistic Critica For right tail test e is calculated using p (TC) = X to i

Please note :occal PAGE x 1000 1000-re Now, . TE X 1000 0.5 0.5 = 3 DOOC 3 x=000 C de OR since xn Binomial (10oo, p. - PX < 1000CP 2015 E Cx) = 1000p. VOX) = 1000 PCI-P) coToooc a - 1 - 1000-20 0.5 0.5 E x=0 under Ho, = Central Usina limit E CX) tovcx) theorem. 1000 X 0.5 = 500 1000 x0.5 x0.5 = 250 - X INCO,1) 2) PCT >cl p=0.5) = p [ioco dc | P=0-57 = p [ x > looo cl T - ECT VCT) Here T- E (1 800) X > 1000 C

under Null hypothesis. P (TOC) = P(Z > C-0.5 ſ0.00025 E CT) = 0.5 v (1) = 0,9 x OS = 0.00025 using for this Normal probability table given we can calutate probability. ". ~ Nco, i). I-0.5 70.00025 Please Note Dear student Generally for part 4 and part bo we take p (T>) - . assuming Ho is true. where x = oil or 0.05 or 0.01 known as significance level. - EGT) & cror | VCN No-001 P(T>c) = PCE-E CT) 2.(-0.57 CT) 2 C-0.5 c Here X: 560 het 2= T- ECT) O NCO,U). VUCT) TL 560 = 0.56. 1000 =

Now under Ho is true = z = T-ECT) svet) 0,56 -0.5 0.00025 T ~ Ncoil o z = 3.794733192 - Critical region is 1.645 ice_ > 3.794733192 0 1•645 in he reject the bull bypothesis