39. It is desired to test the hypothesis H0: θ = (1)/(3) against H1: θ = (2)/(3) on a Bernoulli

Step 1: Under $H_0$, the probability of accepting $H_0$ is $frac{1}{3}$ and the probability of

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39. It is desired to test the hypothesis $H_0: \theta = \frac{1}{3}$ against $H_1: \theta = \frac{2}{3}$ on a Bernoulli random variable $X$ with $P(X = 1) = \theta = 1 - P(X = 0)K$, sequentially. The procedure adopted is to accept $H_0$ after the $i^{th}$ observation $X_i$ if $X_i - X_{i-1} = -1$, to accept $H_1$ if $X_i - X_{i-1} = 1$ and to continue sampling otherwise. Let $N$ denote to reach a decision. then the distribution of $N$ under $H_0$ is:-\\ (a) $\frac{2 + 2^{n-1}}{3^n}$ \quad (b) $\frac{2 + 2^{n-2}}{3^{n-1}}$ \quad (c) $\frac{2 + 2^{n-4}}{3^{n-2}}$ \quad (d) None of these

          39. It is desired to test the hypothesis $H_0: \theta = \frac{1}{3}$ against $H_1: \theta = \frac{2}{3}$ on a Bernoulli random variable $X$ with $P(X = 1) = \theta = 1 - P(X = 0)K$, sequentially. The procedure adopted is to accept $H_0$ after the $i^{th}$ observation $X_i$ if $X_i - X_{i-1} = -1$, to accept $H_1$ if $X_i - X_{i-1} = 1$ and to continue sampling otherwise. Let $N$ denote to reach a decision. then the distribution of $N$ under $H_0$ is:-\\
(a) $\frac{2 + 2^{n-1}}{3^n}$ \quad (b) $\frac{2 + 2^{n-2}}{3^{n-1}}$ \quad (c) $\frac{2 + 2^{n-4}}{3^{n-2}}$ \quad (d) None of these

39. It is desired to test the hypothesis H0: θ = (1)/(3) against H1: θ = (2)/(3) on a Bernoulli random variable X with P(X = 1) = θ = 1 - P(X = 0)K, sequentially. The procedure adopted is to accept H0 after the i^th observation Xi if Xi - Xi-1 = -1, to accept H1 if Xi - Xi-1 = 1 and to continue sampling otherwise. Let N denote to reach a decision. then the distribution of N under H0 is:-

(a) (2 + 2^n-1)/(3^n) (b) (2 + 2^n-2)/(3^n-1) (c) (2 + 2^n-4)/(3^n-2) (d) None of these

Added by Sean S.

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