Choosing H0 and Ha in hypothesis testing
Someone explain the difference between Ho: and Ha: and what are they implemented to find?
Question about choosing null vs alternative hypotheses in hypothesis testing
Null hypothesis and Alternative Hypothesis
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Your null hypothesis is $H_0:p=0.3$
The alternative hypothesis is $H_1:p>0.3$
You need to calculate $$p(X\geq317)$$ using $X\sim Bin(1000,0.3)$
Can you finish?
Just to clarify:
- The null hypothesis always has an equal sign and never an inequality symbol
- In this particular example we conclude that $317$ is not in the critical region.
We conclude that in accepting the null hypothesis there is insufficient evidence that the probability is more than $30$%
Both ideas of the null and alternative hypothesis are true. The null hypothesis must always include an equals sign, whether it be $\geq\text{, } \leq\text{, or just}=$. Usually, however, it's just $=$. The alternative hypothesis is what we wish to show.
The null hypothesis in this case is that the proportion of children in economically disadvantaged areas raised in single-parent homes is $30$%.
The alternative hypothesis is that the proportion of children in economically disadvantaged areas raised in single-parent homes is greater than $30$%.
More formally
$$H_0 : p=0.3$$
$$H_a : p \gt 0.3$$
There are two ways you can test this hypothesis if you so wish. Letting $X$ be the number of children raised in single-parent homes, you can use normal approximation to the binomial:
$$P(X\geq317)=1-P(X\lt317)=1-\Phi\left(\frac{316.5-300}{\sqrt{1000\cdot0.3\cdot0.7}}\right)$$
where I used a continuity correction
In R statistical software
> 1-pnorm((316.5-300)/sqrt(1000*.3*.7))
[1] 0.1274333
You could also, using software, find the exact probability using the standard binomial distribution:
$$P(X\geq317)=\sum_{k=317}^{1000} {1000 \choose k}\cdot0.3^k\cdot0.7^{1000-k}$$
> sum(dbinom(317:1000,1000,.3))
[1] 0.1277011
Since $n$ is large, the normal approximation does very well.
At $\alpha=0.05$ we fail to reject the null hypothesis.