In: Statistics and Probability
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. Typical rain is somewhat acidic, so acid rain is defined as rainfall with pH below 5.0. Suppose that pH measurements of rainfall on different days in a Canadian forest follow a Normal distribution with standard deviation σ=0.5. A researcher wants to determine if there is good evidence that the mean pH μ for all rainy days is less than 5.0. The researcher has collected a sample of size n=15 with sample mean being x ̅=4.8.
The researcher wants to determine if there is good evidence that the mean pH μ for all rainy days is less than 5.0. The researcher has collected a sample of size n=15 with a sample mean being x ̅=4.8 and the population standard deviation σ=0.5.
So, based on the given researchers belief the hypotheses are:
a) The Null hypothesis
b) The Alternate hypothesis
c) The mean and the standard deviation of the sampling distribution is
Mean = = 4.8 and the standard deviation of the sampling distribution is computed as:
d) To calculate the P-value we need to the test statistic that is calculated as:
P-value:
The P-value is computed using the excel formula for normal distribution which is =NORM.S.DIST(-1.55, TRUE), thus the P-value is 0.0606.
e) Decision:
Since P-value is greater than 0.05 hence failed to reject the null hypothesis.
Conclusion:
Since we cannot reject the null hypothesis hence we conclude that there is insufficient evidence to support the claim that the mean pH μ for all rainy days is less than 5.0.