In: Statistics and Probability
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
Assume a catfall is equally likely to occur in each month and carry out a suitable test.
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.
Which method out of (a) and (b) would be best to use, if it were reasonable to assume that a cat could fall at any moment? Explain, briefly.
a) Here we define
Null Hypothesis: Rate of cat falling does not varies between months of the year
Alternate Hypothesis : Rate of cat falling varies between months of the year
at 5% level of significance we use the chi square statistics
Months | Number of cat falling | Expected no of cat falling | |||
Observed (1) | Expected (2) | O-E (3) | (O-E)^2 (4) | (4) /(2) | |
Jan | 4 | 9.92 | -5.92 | 35.01 | 3.53 |
Feb | 6 | 9.92 | -3.92 | 15.34 | 1.55 |
Mar | 8 | 9.92 | -1.92 | 3.67 | 0.37 |
Apr | 10 | 9.92 | 0.08 | 0.01 | 0.00 |
May | 9 | 9.92 | -0.92 | 0.84 | 0.08 |
Jun | 14 | 9.92 | 4.08 | 16.67 | 1.68 |
Jul | 19 | 9.92 | 9.08 | 82.51 | 8.32 |
Aug | 13 | 9.92 | 3.08 | 9.51 | 0.96 |
Sep | 12 | 9.92 | 2.08 | 4.34 | 0.44 |
Oct | 12 | 9.92 | 2.08 | 4.34 | 0.44 |
Nov | 7 | 9.92 | -2.92 | 8.51 | 0.86 |
Dec | 5 | 9.92 | -4.92 | 24.17 | 2.44 |
Total | 119 | 20.6639 |
at 5 % level of significance the tabulated value of Chi Square is 19.7 at 11 degree of freedom
since calculated value of chi square ( 20.66) is greater than tabulated value ( 19.7)
Therefore Null Hypothesis is rejected and we conclude that rate of cat falling varies between the months
b)
Null Hypothesis: Rate of cat falling does not varies between days of the year
Alternate Hypothesis : Rate of cat falling varies between Days of the year
at 5% level of significance we use the chi square statistics
Months | Number of cat falling | Number of days | Cat fallen per day | ||||
Observed (1) | observed | expected | O-E | (O-E)^2 | (O-E)^2 /E | ||
Jan | 4 | 31 | 0.13 | 0.33 | -0.20 | 0.040 | 0.122 |
Feb | 6 | 28 | 0.21 | 0.33 | -0.12 | 0.013 | 0.041 |
Mar | 8 | 31 | 0.26 | 0.33 | -0.07 | 0.005 | 0.016 |
Apr | 10 | 30 | 0.33 | 0.33 | 0.00 | 0.000 | 0.000 |
May | 9 | 31 | 0.29 | 0.33 | -0.04 | 0.002 | 0.005 |
Jun | 14 | 30 | 0.47 | 0.33 | 0.14 | 0.019 | 0.057 |
Jul | 19 | 31 | 0.61 | 0.33 | 0.28 | 0.080 | 0.243 |
Aug | 13 | 31 | 0.42 | 0.33 | 0.09 | 0.008 | 0.024 |
Sep | 12 | 30 | 0.40 | 0.33 | 0.07 | 0.005 | 0.015 |
Oct | 12 | 31 | 0.39 | 0.33 | 0.06 | 0.003 | 0.010 |
Nov | 7 | 30 | 0.23 | 0.33 | -0.10 | 0.009 | 0.028 |
Dec | 5 | 31 | 0.16 | 0.33 | -0.17 | 0.028 | 0.086 |
Total | 119 | 365 | 0.646 | ||||
Expected | 365/119 |
here for 365-1 = 364 degree of freedom the calculated value of chi
square is 0.646 which is very much less than the tabulated
value
therefore, we do not reject the null hypothesis and conclude that rate of cat falling does not varies between days
c) the method on the basis of days is better to chose to assume the cat can fall at any moment as it is independent if days and so of moment.
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