Answer:
1.-Then we p-value indicates that we are in the rejection region we reject H₀
We support the claim of the garbage collector the average picking up more than 4 tn of garbage
2.- p = 75 %
3.-3.-t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim
Step-by-step explanation:
1.- If p-value is 0,04 and significance level α = 5 % or α = 0,05 then p-value < α
Test hypothesis should be ( x the average of garbage)
Null hypothesis H₀ x = 4 Tn
Alternative Hypothesis Hₐ x > 4 Tn
Then alternative hypothesis suggests a one tail-test to the right and
p-value < 0,05
Then we p-value indicates that we are in the rejection region we reject H₀
We support the claim of the garbage collector the average picking up more than 4 tn of garbage
2.- As we are dealing with a normal distribution the CI 95 % is symmetrical with respect to the mean, therefore the proportion of student living in Brooklyn is:
(0,73 + 0,77) /2
p = 0,75 p = 75 %
Test hypothesis:
Null Hypothesis H₀ μ = 26
Alternative Hypothesis Hₐ μ < 26
Alternative hypothesis tells us the test is a one-tail test to the left
Sample size n = 25
Sample mean μ = 24,4
Sample Standard deviation = 9,2
We assume normal distribution, and as n < 30 we use t-student table
with 24 degree of freedom
Significance level is 0,05 and df = 24 we find t (c) in t- student table
t(c) = 1,7109 test to the left t(c) = -1,7109
To calculate t(s)
t(s) = ( 24,4 - 26 ) / 9,2 / √25
t(s) = - 1,6 * 5 / 9,2
t(s) = - 0,87
Comparing t(s) and t(c) we have
t(s) > t(c)
t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim
