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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?

Respuesta :

Answer:

Case n =5

[tex]z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894[/tex]  

Case n =15

[tex]z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549[/tex]  

Case n = 40

[tex]z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530[/tex]  

P value

Case n =5

[tex]p_v =P(z<-0.894)=0.186[/tex]  

Case n =15

[tex]p_v =P(z<-1.549)=0.0606[/tex]  

Case n =40

[tex]p_v =P(z<-2.530)=0.0057[/tex]  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=4.8[/tex] represent the sample mean

[tex]\sigma=0.5[/tex] represent the population standard deviation

[tex]n[/tex] sample size  

[tex]\mu_o =5[/tex] represent the value that we want to test  

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] 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:[tex]\mu \geq 5[/tex]  

Alternative hypothesis:[tex]\mu < 5[/tex]  

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:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (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

[tex]z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894[/tex]  

Case n =15

[tex]z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549[/tex]  

Case n = 40

[tex]z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530[/tex]  

P-value  

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

Case n =5

[tex]p_v =P(z<-0.894)=0.186[/tex]  

Case n =15

[tex]p_v =P(z<-1.549)=0.0606[/tex]  

Case n =40

[tex]p_v =P(z<-2.530)=0.0057[/tex]  

Using the z-distribution, it is found that:

For n = 5, the p-value is of 0.1867.

For n = 15, the p-value is of 0.0606.

For n = 40, the p-value is of 0.0057.

At the null hypothesis, it is tested if the pH is of 5, that is:

[tex]H_0: \mu = 5[/tex]

At the alternative hypothesis, it is tested if the pH is of less than 5, that is:

[tex]H_1: \mu < 5[/tex]

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • [tex]\sigma[/tex] is the standard deviation of the sample.
  • n is the sample size.

For this problem, the values of the parameters are: [tex]\overline{x} = 4.8, \mu = 5, \sigma = 0.5[/tex]

For a sample of 5, we have that [tex]n = 5[/tex], and hence:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{4.8 - 5}{\frac{0.5}{\sqrt{5}}}[/tex]

[tex]z = -0.89[/tex]

The p-value is the probability of finding a sample mean pH below 4.8, which is the p-value of z = -0.89.

Looking at the z-table, z = -0.89 has a p-value of 0.1867, hence, the p-value of the test is of 0.1867.

For a sample of 15, we have that [tex]n = 15[/tex], and hence:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{4.8 - 5}{\frac{0.5}{\sqrt{15}}}[/tex]

[tex]z = -1.55[/tex]

[tex]z = -1.55[/tex] has a p-value of 0.0606, hence, this is the p-value of the test.

For a sample of 40, we have that [tex]n = 40[/tex], and hence:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{4.8 - 5}{\frac{0.5}{\sqrt{40}}}[/tex]

[tex]z = -2.53[/tex]

[tex]z = -2.53[/tex] has a p-value of 0.0057, hence, this is the p-value of the test.

A similar problem is given at https://brainly.com/question/24146681

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