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 results 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 Washington forest follow a Normal distribution with standard deviation sigma = 0.5. A sample of n days finds that the mean pH is x^- = 4.8. Is this good evidence that the mean pH mu p for all rainy days is less than 5.0? The answer depends on the size of the sample. In all cases, use: sigma = 0.5 x^- = 4.8 in all three tests alpha = 0.05 a) Conduct a hypothesis test with n = 5 b) Conduct a hypothesis test with n = 15 c) Conduct a hypothesis test with n = 40 For each test, your answer should include: hypotheses, a test statistic, a p-value, and a conclusion.

Explanation / Answer

null hypothesis: mean=5

alternate hypothesis: mean<5

a) for n=5; std error =std deviation/(n)1/2 =0.2236

test stat z=(X-mean)/std error =(4.8-5)/0.2236=-0.8944

for above p value =0.1855

as p value is above 0.05 we can not reject null hypothesis.

we do not have sufficient evidence to conclude that mean pH value for all rainy days is less then 5.

b)

for n=15; std error =std deviation/(n)1/2 =0.1291

test stat z=(X-mean)/std error =(4.8-5)/0.1291=-1.5492

for above p value =0.0607

as p value is above 0.05 we can not reject null hypothesis.

we do not have sufficient evidence to conclude that mean pH value for all rainy days is less then 5.

c(

for n=40; std error =std deviation/(n)1/2 =0.0791

test stat z=(X-mean)/std error =(4.8-5)/0.0791=-2.5298

for above p value =0.0057

as p value is less then 0.05 we can reject null hypothesis.

we have sufficient evidence to conclude that mean pH value for all rainy days is less then 5.

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