To make a reasoned opinion on whether or not the die is unbiased, we can turn to a Chi Squared Goodness of Fit test.
We will start by making a null hypothesis and an alternate hypothesis. Our null hypothesis #H_0# can be the statement "The die is unbiased and all sides are equally likely to be rolled." One possible alternate hypothesis #H_A# can be "The die is biased and one or more sides are more likely to be rolled than other sides."
We start by considering the options for our die. In this problem, there are two options:
Event #A# = A throw of 3 or 4 happens
Event #B# = A throw of 3 or 4 does not happen
For a fair unbiased die with six sides, the probability of any particular side being rolled is #1/6#. Thus, we can assign probabilities to these two events:
#P(A) = 2/6 = 1/3#
#P(B) = 4/6 = 2/3#
To conduct the #chi^2# test, we need to determine the number of occurrences we observed, along with the number of occurrences we expected under our #H_0#. The observed is easy, as it was given to us: 3240 instances of event A, and (by implication) 5760 instances of event B.
The expected number of event A instances can be determined by multiplying the number of attempts by the probability of the event. Thus: #1/3 * 9000 = 3000#. Thus, the expected number of event B instances would be 6000.
We have the information we need to determine the #chi^2# value for our test. We refer to the #chi^2# formula:
#chi^2 = sum_"i=1"^n ((O_i-E_i)^2)/E_i#
where #n# is the number of events we are tracking, #O_i# is the observed count of event number #i# happening, and #E_i# is the expected count of event number #i# happening. The #sum# symbol means we calculated these values for all events, and add them together.
Here's a table to illustrate:
#[("Event", "|" , O_i - E_i, "|" , (O_i - E_i)^2, "|" , (O_i - E_i)^2/E_i),
(A, "|", 240, "|", 57600, "|", 19.2),
(B, "|", -240, "|", 57600, "|", 9.6),
("", "", "", "", "Sum", "|", 28.8) ]#
The #chi^2# value we have for our test is 28.8. To interpret this in relation to our #H_0#, we need to consult a #chi^2# table to determine what the #chi^2# critical value is for our test. Since we have 2 events we are tracking, we use one less than that (aka 1) as our degrees of freedom (#df = 1#). For an #alpha = 0.05# level of significance with #df = 1#, a #chi^2# table gives a critical value of 3.841.
Since our #chi^2 = 28.8 > 3.841#, we can say that at the 5% level of significance, enough statistical weight exists for us to reject our null hypothesis #H_0# that the die is unbiased. Although this does not mean we can "accept" the alternate hypothesis, it does mean we have a strong confidence that the die is, in fact, biased. Keep in mind that a 5% significance level does not rule out that the die truly is unbiased; one way to look at it is there is a less than 5% chance that the die is unbiased and the results we saw in our experiment were due to random variation in the rolling.
