Question

The observed value of a random sample of size \(9\) from a distribution having continuous and strictly increasing cumulative distribution function is as below:

Hint:

In a one-sided sign test for \(H_1:M>0\), count the number of positive signs and calculate the upper-tail binomial probability under \(H_0\).

Answer 1

Correct Answer: 1.5

Step 1: Count positive and negative observations.
The positive observations are
\[ 1.00,\;0.70,\;0.25,\;1.20,\;0.68 \] So, the number of positive signs is
\[ S=5 \] There are \(4\) negative observations.

Step 2: Use the sign test under \(H_0\).
Under \(H_0:M=0\), the number of positive signs follows
\[ S\sim Binomial\left(9,\frac12\right) \] Since the alternative is
\[ H_1:M>0, \] large values of \(S\) support \(H_1\).
Therefore, the \(p\)-value is
\[ p=P(S\geq 5) \]

Step 3: Calculate the \(p\)-value.
\[ p=P(S\geq 5) = \sum_{k=5}^{9}\binom{9}{k}\left(\frac12\right)^9 \] By symmetry of the binomial distribution with \(p=\frac12\),
\[ P(S\geq 5)=\frac12 \] Thus,
\[ p=0.50 \]

Step 4: Decide whether \(H_0\) is rejected.
At level \(0.05\), since
\[ p=0.50>0.05, \] we do not reject \(H_0\).
Therefore, \(H_0\) is accepted and
\[ \eta=1 \]

Step 5: Find \(p+\eta\).
\[ p+\eta=0.50+1 \] \[ =1.50 \]

Step 6: Final conclusion.
Hence,
\[ \boxed{1.50} \]