Question

An inspector for the state gaming commission visited a casino to inspect a card-dealing machine. The inspector observed the suits of a small sample of the cards in the machine, and counted 404 spades, 430 hearts, 400 diamonds, and 366 clubs.

1- Calculate the Chi-squared value based on the assumption that the cards represent a random assortment.

2- Are the observed cards consistent with a random assortment of cards, or are the discrepancies too much to be random?

I need some help in showing the method of answering this question.

Answer 1

The observed counts of suits are

Suit	Observed Frequency (fo)
Spades	404
Hearts	430
Diamonds	400
Clubs	366
Total	1600

We know that each of these 4 suits are equally probable in a random assortment of cards. That is the probability of a randomly selected card is of each of these types is 1/4. The inspector is interested in knowing if the cards represent the random assortment, that is there is an equal probability of selecting any of the suits.

We want to test the following hypotheses.

$\begin{align*} &H_0: \text{The cards represent a random assortment}\leftarrow\text{null hypothesis}\\ &H_a: \text{The cards do not represent a random assortment}\leftarrow\text{alternative hypothesis}\\ &\alpha=0.05\leftarrow\text{level of significance to test the hypotheses} \end{align*}$

If the cards represent a random assortment, then among 1600 total cards sampled by the inspector, the expected frequency of each suit would be 1600/4=400

the inspector would have expected an $\begin{align*} f_e=400 \end{align*}$ of each suit in the sample.

Suit	Observed Frequency (fo)	Expected Frequency (fe)
Spades	404	400
Hearts	430	400
Diamonds	400	400
Clubs	366	400

The chi-square test statistics is calculated as

$\begin{align*} \chi^2&=\sum\frac{(f_o-f_e)^2}{f_e}\\ &=\frac{(404-400)^2}{400}+ \frac{(430-400)^2}{400}+ \frac{(400-400)^2}{400}+ \frac{(366-400)^2}{400}\\ &=5.18 \end{align*}$

The degrees of freedom for chi-square is k-1=4-1=3, where k=4 is the number of groups.

The critical value for alpha=0.05 using the chi-square table for df=3 is 7.815.

We will reject the null hypothesis if the test statistics is greater than the critical value.

Here the test statistics of 5.18 is less than the critical value of 7.815.

That means we will not reject the null hypothesis.

We can conclude that the cards represent a random assortment and the discrepancies can be attributed to chance.