Question 2 Thirty samples were taken from a stream, and the pollution level in p

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Question 2 Thirty samples were taken from a stream, and the pollution level in parts per million (ppm) was recorded for each sample. The average pollution level for the data was T 10.1ppm. Suppose that the population standard deviation is 2.7ppm. The investigator who collected the samples is interested in determining whether or not there is evidence that the population mean pollution level Au is greater than 9.5ppm. The hypotheses are Ho Au 9.5 and Ha Au 9.5. (a) Given a 0.05, for what values of z will you reject the null hypothesis (b) Using o 2.7 and w 9.5, for what values of will you reject Ho at a 0.05? (c) Using o F 2.7 and pu 10, what is the probability that T will fall in the region defined in part (b)? This probability is the power of the test when Au 10 (d) Find the power of the test when Au 11 and compare it with your answer in part (c

Explanation / Answer

let X be the random variable denoting the pollution level in ppm.

the population standard deviation is given as 2.7 ppm.

assumption is that X follows a normal distribution with unknown mean u.

hence X~N(u,2.72)

and we are to test

H0: u=9.5   vs H1: u>9.5

to test this we have a random sample of size n=30 with sample mean Xbar=10.1 ppm

now Xbar~N(u,2.72/30)

a) hence Z=(Xbar-u)*sqrt(30)/2.7~N(0,1) distribution

so for given level of significance alpha=0.05 H0 will be rejected iff

z>taoalpha   where z is the observed value of Z and taoalpha is the upper alpha point of a N(0,1) distribution.

for alpha=0.05 tao0.05=1.64

hence for the values of Z which are greater than 1.64 we reject the null hypothesis.

b) so from a) we reject H0 at alpha=0.05 when

Z>1.64

or, (Xbar-u)*sqrt(30)/2.7>1.64

or, Xbar>u+1.64*sqrt(30)/2.7

so for u=9.5

Xbar>9.5+1.64*sqrt(30)/2.7=12.8

hence for Xbar values greater than 12.8 ppm H0 will be rejected at alpha=0.05 [answer]

c) so when u=10

Xbar~N(10,2.72/30)

so the probability that Xbar will fall in the region as mentioned in b) is

P[Xbar>12.8]

=P[(Xbar-10)*sqrt(30)/2.7>(12.8-10)*sqrt(30)/2.7]

=P[Z>5.68]=1-P[Z<5.68]=1-1=0   [using MINITAB]

this is the power of the test when u=10

d) hence power of the test when u=11 is

P[(Xbar-11)*sqrt(30)/2.7>(12.8-11)*sqrt(30)/2.7]

=P[Z>3.65]=1-P[Z<3.65]=1-0.999869=0.000131 [answer] [using MINITAB]

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