# Z-test for a population proportion

I'm asked to carry out a Z-test for a population proportion on whether a die is fair. The only information I'm given is that "A die fell showing a "four" on $124$ occasions out of $500$".

I'm not sure how to do this, but after giving it a try, this is what I have:

$H_0: p = 0.1667$ (die is fair, outcomes are equally likely)

$H_1: p \neq 0.1667$

Here, I will assume that the significance level is $\alpha = 5\%$

Using the Z-test statistic formula, I have that $Z_{\text{observed}} = 10.9$

Is this right? Would really appreciate some help.

• Which formula are you using for your calculation. As a hint, you are testing a single proportion against an assumed value. – mfloren Mar 5 '17 at 22:54
• @mfloren hi, I added a picture of the formula into the question. Regarding your hint, so do I just do it like a normal Z-test? – NDZS Mar 5 '17 at 23:37

If the die is fair, the proportion of $4$'s to expect is $1/6=0.1667$. Is your observed proportion of $p_{\text{ obs}}=124/500=0.248$ far enough from it to exclude the null hypothesis, $H_0: p_{\text{ true}}=0.1667$, based on the observed proportion at a $5\%$ significance level?

You can investigate this in two different ways:

1. Using asymptotic calculations (normal approximation or Z test):

$$\text{test statistic}= \frac{p_{\text{ obs}} - p_{\text{ true}}}{\sqrt{\frac{p_{\text{ true}}\,(1\,-\,p_{\text{ true}})}{n}}}=\frac{0.248-0.1667}{\sqrt{\frac{0.1667\times(1-0.1667)}{500}}}=4.88$$

1. Using the binomial test:

$$\Pr(\text{fours} \geq 124)= \sum_{x=124}^{500} {500\choose x} \,0.1667^x\, (1-0.1667)^{500-x}$$

a calculation that explains the reliance on the first method (particularly before modern computers).