Emissions of sulfur dioxide by industry set off chemical changes in the atmosphe

Question

Emissions of sulfur dioxide by industry set off chemical changes in the atmosphere that result in "acid rain." The acidity of liquids is measured by pH on a scale of 0 to 14. Distilled water has pH 7.0, and lower pH values indicate acidity. Normal rain is somewhat acidic, so acid rain is sometimes defined as rainfall with a pH below 5.0. Suppose that pH measurements of rainfall on different days in a Canadian forest follow a Normal distribution with standard deviation = 0.5. You plan to test the following hypotheses at the 5% level of significance.
H0: = 5
Ha: < 5
You want to use a test that will almost always reject H0 when the true mean pH is 4.6. Find the power of the test for a sample of size n = 18 by following these steps.
(a) Write the z statistic for a sample of size 18. What values of z lead to rejecting H0 at the 5% significance level? (Round your answer to two decimal places.)
z <

(b) Starting from your result in part (a), what values of x lead to rejecting H0? (Round your answer to one decimal place.)
x <

(c) What is the probability of rejecting H0 when = 4.6? This probability is the power against this alternative. (Round your answer to four decimal places.)

Explanation / Answer

Test Used: Z-Test For Single Mean
Set Up Hypothesis
Null, defined as rainfall with equals 5.0 H0: U=5
Alternate, defined as rainfall with a pH below 5.0 H1: U<5
Test Statistic
Population Mean(U)=5
Given That X(Mean)=4.6
Standard Deviation(S.D)=0.5
Number (n)=16
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=4.6-5/(0.5/Sqrt(16)
Zo =-3.2
| Zo | =3.2
Critical Value
The Value of |Z | at LOS 0.05% is 1.64
We got |Zo| =3.2 & | Z | =1.64
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value : Left Tail - Ha : ( P < -3.2 ) = 0.0007
Hence Value of P0.05 > 0.0007, Here we Reject Ho

a)Zo =-3.2 b) X<-1.64

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