Pls help me to do the following
Q)Carryout the following test using the binomial distribution where random variable X represents the number of successes.
H0: p = 0.50; H1: p ¹ 0.05; n = 20, x = 7 and using a 10% level of significance.
Suppose that $\displaystyle H0$ holds, then calculate the probability of getting a result outside the interval $\displaystyle np\pm 2$, with $\displaystyle p=0.5$. If this probability falls below $\displaystyle 0.1$ then you reject the null hypothesis at the $\displaystyle 10\%$ level of significance as the observed result is outside the given interval.
You calculate the probability using the binomial distribution:
$\displaystyle P(x \in (8,12))=\sum_{i=8}^{12} b(i,0.5,20)$
where $\displaystyle b(i,p,20)$ is the probability of exactly $\displaystyle i$ success in $\displaystyle n$ trials with probability of success on a single trial $\displaystyle p$ for the binomial distribution.
CB
Thanks captain.
I figured it in the following way. Pls say whether this is alright.
Test static X has the binomial distribution which can be written as
X ~B(20,0.5)
This seems to be a two tailed test and since the value
of p given for H0 is 0.5, then the distribution of the test static X will be symmetrical.
From the table of binomial distribution
P(x<=7)=0.1316
So the probability for the two tailed test will be = 0.1316*2=0.2632
Hence we cannot reject H0 for the significance level of 10%.
Is this argument correct? Pls comment.