Question

Emissions of sulfur dioxide by industry sett 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 σ = 0.5. A sample of n days finds that the mean pH is x¯¯¯ = 4.8. What are the P- Values for tests of sample sizes n = 5, n = 15 and n = 40?

Answer 1

Answer:

Case n =5

$z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894$  

Case n =15

$z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549$  

Case n = 40

$z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530$  

P value

Case n =5

$p_v =P(z$  

Case n =15

$p_v =P(z$  

Case n =40

$p_v =P(z$  

Step-by-step explanation:

Data given and notation  

$\bar X=4.8$ represent the sample mean

$\sigma=0.5$ represent the population standard deviation

$n$ sample size  

$\mu_o =5$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is lower than 5, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 5$  

Alternative hypothesis:$\mu < 5$  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

Case n =5

$z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894$  

Case n =15

$z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549$  

Case n = 40

$z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530$  

P-value  

Since is a left tailed test the p value would be:  

Case n =5

$p_v =P(z$  

Case n =15

$p_v =P(z$  

Case n =40

$p_v =P(z$