Question

The restaurant owner Lobster Jack wants to find out what the peak demand periods are, during the hours of operation, in order to be better prepared to serve his customers. He thinks that, on average, 60% of the daily customers come between 6:00pm and 8:59pm (equally distributed in that time) and the remaining 40% of customers come at other times during the operating hours (again equally distributed). He wants to verify if that is true or not, so he asked his staff to write down during one week the number of customers that come into the restaurant at a given hour each day. His staff gave him the following data:

Time	Day 1	Day 2	Day 3	Day 4	Day 5	Day 6	Day 7
5:00pm-5:59pm	15	19	21	20	12	15	15
6:00pm-6:59pm	30	23	24	25	28	29	26
7:00pm-7:59pm	36	29	39	35	39	30	32
8:00pm-8:59pm	29	33	23	29	24	32	27
9:00pm-9:59pm	21	20	12	19	18	14	20
10:00pm-10:59pm	12	12	15	12	10	15	14
11:00pm-11:59pm	8	7	9	10	12	12	9

Help the manager figure out if his instincts are correct or not. Use a Chi-Squared test to see if the observed distribution is similar to the expected. Use the average demand for a given time as your observed value.

Answer 1

Let our two Hypothesis be at 5 % significance level :

Null Hypothesis :H(0) = Observed = Expected

Alternate Hypothesis: H(1) = Observed ≠ Expected

	Day
Time Interval	1	2	3	4	5	6	7	Average
5:00pm-5:59pm	15	19	21	20	12	15	15	16.71429
6:00pm-6:59pm	30	23	24	25	28	29	26	26.42857
7:00pm-7:59pm	36	29	39	35	39	30	32	34.28571
8:00pm-8:59pm	29	33	23	29	24	32	27	28.14286
9:00pm-9:59pm	21	20	12	19	18	14	20	17.71429
10:00pm-10:59pm	12	12	15	12	10	15	14	12.85714
11:00pm-11:59pm	8	7	9	10	12	12	9	9.571429

	6:00pm-8:59pm	Rest of the Day
Percentage	60%	40%
Observed	88.85714	56.85714
Expected	87.42857143	58.28571429
O-E	1.428568571	-1.428574286
(O-E)^2	2.040808163	2.04082449
(O-E)^2/E	0.023342577	0.035014146
Summation((O-E)^2/E) =	0.058356723
	Chi-Squared Statistic =	Summation((O-E)^2/E)

Since the number of values is 2 the degrees of freedom is n- 1= 2-1 = 1

and the level of significance is at 5 %;

according to the chi-squared chart at of = 1 and 5 % level of significance, the critical value is = 3.84 ;

Since our chi-squared value i.e 0.058356723 is lesser than the critical value, we accept the null hypothesis i.e :

Null Hypothesis :H(0) = Observed = Expected

i.e observed values are equal to the expected values.

Hence we can say that the observed distribution is similar to the expected.