Answer:
H0: μ = 5 versus Ha: μ < 5.
Step-by-step explanation:
Given:
μ = true average radioactivity level(picocuries per liter)
5 pCi/L = dividing line between safe and unsafe water
The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.
A type I error, is an error where the null hypothesis, H0 is rejected when it is true.
We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.
Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.
Answer:
A=Both distributions are nearly symmetric
Step-by-step explanation:
To begin lets say year one the population is 2,000
assume this in percent we can say that 2,000=100%
at the end of this year the population will have grown by 4% meaning that
100%+4% will be the new population =104%
but we know that 100%=2,000 so what about 104%=X?
so (104*2000)/100=2,080
So year 1 the population will have grown by 80 people amounting to 2080 people.
Now do it manually year by year until we reach the 15th year. i will help you up to year 3
2nd year= (2080*104)/100=2163.2
3rd year= (2163*104)/100=2249.728
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You can use the calculator it’s easier and plus you don’t have to put four 9’s
