Question

A study of pollutants showed that certain industrial emissions should not exceed 2.3 parts per million. You believe a particular company may be exceeding this average. To test this supposition, you randomly take a sample of nine air tests. The sample average is 3.3 parts per million, with a sample standard deviation of 0.4. Does this result provide enough evidence for you to conclude that the company is exceeding the safe limit? Use α = 0.05. Assume emissions are normally distributed.

Answer 1

Answer:

$t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5$    

The degrees of freedom are given by:

$df=n-1=9-1=8$  

And the p value would be:

$p_v =P(t_{(8)}>7.5)=0.0000346$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

Step-by-step explanation:

Information provided

$\bar X=3.3$ represent the sample mean in ppm

$s=0.4$ represent the sample standard deviation

$n=9$ sample size  

$\mu_o =2.3$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic  

$p_v$ represent the p value

System of hypothesis

We want to verify if the true mean is higher than 2,3 ppm, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 2.3$  

Alternative hypothesis:$\mu > 2.3$  

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the info given we got:

$t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5$    

The degrees of freedom are given by:

$df=n-1=9-1=8$  

And the p value would be:

$p_v =P(t_{(8)}>7.5)=0.0000346$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit