What hypothesis test is suitable in this case?

Question

Null: Event frequency does not vary by weekday Alternate: Event frequency varies by weekday

Data: Day: Mon, Tue, Wed, Thu, Fri, Sat, Sun event_count: 12, 15, 20, 10, 19, 10, 11

What hypothesis test (e.g chi-squared, p-test) would be best to investigate this relationship?

Answer 1

Outline of the appropriate chi-squared test:

Chi-squared goodness-of-fit test. Null hypothesis is that days are equally likely categories for events: probability $1/7$ for each day. So out of your 97 events the 'expected number' on each day is $E = 97/7 = 13.85714.$

Under the null hypothesis, the test statistic $$Q = \sum_{i=1}^7 \frac{(X_i - E)^2}{E}$$ is approximately distributed as $\mathsf{Chisq}(\text{df} = 6).$

Check your text for this formula, possibly with slightly different notation. Also, perhaps you can find a similar problem worked as an example.

The computations below, using R statistical software, show there is no evidence that events are other than evenly distributed across days. You should verify the computations using a calculator and the critical value from a printed table of chi-squared distributions (or using a statistical calculator). [If you use a printed table, look on the row for DF = 6.]

X = c(12, 15, 20, 10, 19, 10, 11);  t = sum(X)
[1] 97
E = 97/7
Q = sum((X - E)^2/E);  Q
[1] 7.71134
qchisq(.95, 6)
[1] 12.59159    # critical value for 5% level test;  Q < critical value
1 - pchisq(Q, 6)
[1] 0.2600232   # P-value of test > 5%

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Note: The histogram below shows simulated values of $Q$ for 100,000 weeks where days are equally likely. The density curve of $\mathsf{Chisq}(6)$ is shown in black. The histogram is well approximated by the chi-squared density.

Here is the approximate distribution of $Q$ when days are unlikely in such a way that two days (perhaps on weekends) share half of the events and the other five days share the other half. [Specifically, alternative probability vector $p_a = (.25, .1, .1, .1, .1, .1, .25).]$ The distribution is nowhere near $\mathsf{Chisq}(6).$ Most of the probability is above $12.6.$

Note: Here is R code for the first figure:

set.seed(1124);  m = 10^5;  n = 97;  E = 97/7; q = numeric(m)
for(i in 1:m) {
  d = sample(1:7, n, rep=T)
  rle.inf = rle(sort(d));  X = rle.inf$len
  q[i] = sum((X-E)^2/E)  }
  hist(q, prob=T, br=50, col="skyblue2", ylim=c(0, .14), 
    main="Simulated CHISQ(6) with Density")
curve(dchisq(x, 6), 0, 25, add=T, lwd=2)