Question

Use the following information to answer the questions. A more recent study of Feline High-Rise Syndrome (FHRS) included data on the month in which each of 119 cats fell ( Vnuk et al. 2004) The data are in the accompanying table. Can we infer that the rate of cat falling varies between months of the year?

Month Number fallen

January 4

February 6

March 8

April 10

May 9

June 14

July 19

August 13

September 12

October 12

November 7

December 5

1. Assume a catfall is equally likely to occur in each month and carry out a suitable test.

2. Assume a catfall is equally likely to occur on each day and carry out a suitable test. Use a calendar for a non-leap year to obtain the needed probabilities.

3. Which method out of (1.) and (2.) would be best to use, if it were reasonable to assume that a cat could fall at any moment? Explain, briefly.

Answer 1

We have to perform chi-square test for goodness of fit.

(1)

We have to test for null hypothesis

against the alternative hypothesis

In case of equally likely occurance, expected frequencies are calculated and comapared with given observed frequencies as follows.

Observed frequency (Oi)	Expected frequency (Ei)	(O-E)2/E
4	9.916667	3.530112
6	9.916667	1.546919
8	9.916667	0.370448
10	9.916667	0.000700
9	9.916667	0.084734
14	9.916667	1.681372
19	9.916667	8.320027
13	9.916667	0.958683
12	9.916667	0.437675
12	9.916667	0.437675
7	9.916667	0.857843
5	9.916667	2.437675
Total 119	119	20.663863

We know,

Here,

number of observations (number of frequencies)

Using calculated values from the table we have

Degrees of freedom

Correspoding [Using R-code '1-pchisq(20.663863,11)']

Though specific level of significance is not mentioned, in practice we take

We reject our null hypothesis if .

Here we observe that .

So, we reject our null hypothesis at 95% confidence level.

Hence, based on given data we can conclude that occurance of catfall is not equally likely in each month.

(2)

We have to test for null hypothesis

against the alternative hypothesis

In case of equally likely occurance, expected frequencies are calculated and comapared with given observed frequencies as follows.

Observed frequency (Oi)	Expected frequency (Ei)    = 119*Number of days in the month / 365	(O-E)2/E
4	10.106850	3.689935
6	9.128767	1.072344
8	10.106850	0.439189
10	9.780822	0.004912
9	10.106850	0.121216
14	9.780822	1.820038
19	10.106850	7.825199
13	10.106850	0.828183
12	9.780822	0.503511
12	10.106850	0.354613
7	9.780822	0.790626
5	10.106850	2.580420
Total 119	119	20.030186

We know,

Here,

number of observations (number of frequencies)

Using calculated values from the table we have

Degrees of freedom

Correspoding [Using R-code '1-pchisq(20.030186,11)']

Though specific level of significance is not mentioned, in practice we take

We reject our null hypothesis if .

Here we observe that .

So, we reject our null hypothesis at 95% confidence level.

Hence, based on given data we can conclude that occurance of catfall is not equally likely in each month.

Comparison-

In a month of more number of days, it is expected of more occurance of catfall and in a month of lesser number of days, it is expected of less occurance of catfall. Thus using the second method is better approch.