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 holds, then calculate the probability of getting a result outside the interval , with . If this probability falls below then you reject the null hypothesis at the level of significance as the observed result is outside the given interval.
You calculate the probability using the binomial distribution:
where is the probability of exactly success in trials with probability of success on a single trial 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.