Question

A veterinary study of horses looked water sources for horses and the investigators found that horses received water from either a well, from city water, or from a stream. The investigators wanted to know if horses are equally likely to get water from each of those three sources. They collected data and observed the following values: Water Source Well City Water Stream Total Observed Count 62 43 31 136 The investigators want to test the hypothesis that horses receive their water equally from a well, from city water or a from a stream.1) Based on this hypothesis what are the expected counts for water source?a. Water Source Well City Water Stream Expected Count 62 43 31 b. Water Source Well City Water Stream Expected Count 0.33 0.33 0.33 c. Water Source Well City Water Stream Expected Count 33 33 33 d. Water Source Well City Water Stream Expected Count 45.33 45.33 45.33 e. Water Source Well City Water Stream Observed Count 33.33 33.33 33.33

Answer 1

Answer:

d. Water Source Well City Water Stream Expected Count 45.33 45.33 45.33

Step-by-step explanation:

Hello!

The objective of this study is to test whether the horses are equally likely to get water from one of three sources (Well, City or Stream), i.e. that the water sources have the same probability of being used.

The study variable is categorical:

X: Origin of the drinking water of the horses, categorized: Well "W", City "C" and Stream "S"

And one study population:

"Horses"

If the study hypothesis is that the three water sources are equally to be selected, you can symbolize it as P(W)=P(C)=P(S)=1/3

You want to test if the observed distribution follows an expected distribution, the test to use is a Chi-Square Goodness of Fit Test.

The hypothesis is:

H₀: P(W)=P(C)=P(S)=1/3

Under the null hypothesis, the formula to calculate the expected frequencies is:

$e_{i}= n * P_{i}$

Where:

$e_{i}$ is the expected value of the i-category.

$P_{i}$ is the probability for the i-category under the null hypothesis.

n= sample size.

Observed frequencies:

Well: 62

City: 43

Stream: 31

Total: 136

Expected frequencies:

$e_{W}= n * P_{W} = 136*1/3 = 45.33$

$e_{C}= n * P_{C}= 136 * 1/3= 45.33$

$e_{S}= n * P_{S}= 136 * 1/3= 45.33$

Typ, if the expected frequencies are correctly calculated, the sum of them must give the total of the sample (although depending on the type of approximation it can vary in one or two decimals)

45.33 * 3= 135.99 ≅ 136

I hope it helps!