A single tree produces about 2.6 × 10² lb of oxygen each year. The Amazon rainforest has about 3.9×1011 trees.
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Answer:
Step-by-step explanation:
Hello!
The objective is to test if the proportion of "X: gloves fitness, categorized: Fit poorly and Fit well" is the same for two populations of interest, "male firefighters" and "female firefighters"
To do this you have to conduct a Chi-Square test of Homogeneity.
In the null hypothesis you have to state that the proportion of the categories of the variable are the same for all the populations of interest.
Be
M: the firefighter is male
F: the firefighter is female
Y: represents the category that the gloves "fit poorly"
W: represents the category that the gloves "fit well"
The null hypothesis will be:
H₀: P(Y|M)=P(Y|F)=P(Y)
P(W|M)=P(W|F)=P(W)
H₁: At least one of the statements in the null hypothesis is false.
α: 0.01
To calculate the statistic under the null hypothesis you have to calculate the expected frequencies first:
O.j= total of the j-column
Oi.= total of the i-row
n= total of observations
r= number of rows (in this case 2)
c=number of columns (in this case 2)
Using the critical value approach, you have to remember that this test is <em><u>always</u></em> one-tailed to the right, meaning that you'll have only one critical value from which the rejection region is defined:
The decision rule is then:
If ≥ 6.635, reject the null hypothesis.
If < 6.635, do not reject the null hypothesis.
The calculated value is greater than the critical value, the decision is to reject the null hypothesis.
So at a 1% level you can conclude that this test is significant. This means that the proportions of gloves fitness, categorized in "Fit poorly" and "Fit well" are different for the male and female firefighters populations.
I hope this helps!
