A biologist is studying the composition of birds on a lake and counts 61 ducks, 17 geese, 11 cranes, 15 swans, and 6 herons. From previous studies performed around the same time of the year, she expects 50% of the birds to be ducks, 23% to be geese, 12% to be cranes, 10% to be swans, and 5% to be herons. What are the correct null and alternative hypotheses for performing a chi-square goodness of fit test

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Mathematics

Hugo Lyons

6 December, 21:46

A biologist is studying the composition of birds on a lake and counts 61 ducks, 17 geese, 11 cranes, 15 swans, and 6 herons. From previous studies performed around the same time of the year, she expects 50% of the birds to be ducks, 23% to be geese, 12% to be cranes, 10% to be swans, and 5% to be herons. What are the correct null and alternative hypotheses for performing a chi-square goodness of fit test

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Allisson Valentine · Answer 1

Observed birds

Number of ducks = 61

Number of geese = 17

Number of cranes = 11

Number of swans = 15

Number of herons = 6

Total number of observed birds = 61 + 17 + 11 + 15 + 6 = 110

Expected birds percentage:

ducks = 50%

geese = 23%

cranes = 12%

swans = 10%

herons = 5%

So, count of expected birds according to given percentage:

ducks = 50/100 * total number of birds = 50/100 * 110 = 55

geese = 23/100 * 110 = 25 (approx)

cranes = 12/100 * 110 = 13 (approx)

swans = 10/100 * 110 = 11 (approx)

herons = 5/100 * 110 = 6 (approx)

The chi-square goodness of fit test is used to compare the observed distribution with the expected distributed distribution of the population. In this,

Null hypothesis: There is no significant difference between observed count and expected count

Alternative hypothesis: There is significant difference between observed count and expected count