1. In 1995, the average level of mercury uptake in wading birds in the Everglades was 15 parts per million. If we assume that the distribution of mercury uptake has a standard deviation of 1 part per million, and a current sample of 25 wading birds has an average of 14.6 parts per million. Which null and alternative hypotheses would be used to test that the level of mercury uptake in the Everglades has decreased since 1995?

We reject H₀, we find that the  current level of mercury uptake in the Everglades has decrease

Step-by-step explanation:

We need to test the current wading bird average 14,6 vs the 1995 average. As sample mean is 14,6 and population mean is 15 we are going to check test is of one tail (left tail test)

Test Hypothesis:

Null hypothesis             H₀          μ   =  μ₀

Alternative hypothesis Hₐ          μ   <  μ₀

Sample size 25             then n < 30 we will use t-student table

Population mean : μ₀ = 15 ppm

sample mean :  μ  = 14,6 ppm

sample standard deviation:   s = 1 ppm

Sample size n :  25

degrees of freedom   df = n - 1         df = 24

We will develop the test for CI 90%  then  α = 10%      α = 0,1   α/2 = 0,05

For that values in t student table we find:

t(c) = 1,711    by symmetry t(c) = - 1,711

To compute ts

t(s) =  ( μ - μ₀ ) / s/√n

t(s) = 14,6 - 15 / 1 /√25

t(s) = - 0,4*5 / 1

t(s) = - 2

Comparing t(c)and t(s)

|t(s)| > |t(c) |      and t(s) < t(c)         -2 < -1,711

Therefore t(s) in in the rejection region, we reject H₀  

step-by-step explanation:

yes it represnts a functio, their are not 2 of the same numbers on the both sides

the answer is d

step-by-step explanation:

you can also use mathway to get you answer + steps