A certain chemical pollutant in the Arkansas River has been constant for several years with mean μ = 34 ppm (parts per million). A group of factory representatives whose companies discharge liquids into the river is now claiming they have lowered the average with improved filtration devices. A group of environmentalists will test to see if this is true. Find the rejection region appropriate for this test if we are using a significance level of 0.05 and have a sample size of 25.

A) Reject H0 if t < -1.960

B) Reject H0 if t < -2.064

C) Reject H0 if t < -1.711

D) Reject H0 if t < -1.708

E) Reject H0 if t < -2.064 or t > 2.064

Question

22-04-2020
Mathematics

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A certain chemical pollutant in the Arkansas River has been constant for several years with mean μ = 34 ppm (parts per million). A group of factory representatives whose companies discharge liquids into the river is now claiming they have lowered the average with improved filtration devices. A group of environmentalists will test to see if this is true. Find the rejection region appropriate for this test if we are using a significance level of 0.05 and have a sample size of 25.

A) Reject H0 if t < -1.960

B) Reject H0 if t < -2.064

C) Reject H0 if t < -1.711

D) Reject H0 if t < -1.708

E) Reject H0 if t < -2.064 or t > 2.064

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Answer:

C) Reject [tex]H_0[/tex] if t < -1.711

Step-by-step explanation:

We are given that a certain chemical pollutant in the Arkansas River has been constant for several years with mean μ = 34 ppm (parts per million).

Also, it is given that the level of significance is 0.05 and a sample size of 25 is taken.

Let [tex]\mu[/tex] = average chemical pollutant in the Arkansas River

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 24 ppm {means that the average is same as before with improved filtration devices}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 24 ppm {means that they have lowered the average with improved filtration devices}

Now, t test statistics is given by; T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, n = sample size = 25

n - 1 = degree of freedom = 25 - 1 = 24

Now, at 0.05 significance level in the t table, the critical value at 24 degree of freedom is given as - 1.711 for left tailed test.

So, we will reject our null hypothesis when the t test statistics is less than the critical value of t which means reject null hypothesis if t < -1.711.