Question

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.1 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 26 samples is 6.6 ppm with a standard deviation of 1.0. Assume the population is normally distributed. A level of significance of 0.01 will be used. Find the value of the test statistic. Round your answer to three decimal places.

Answer 1

Answer:

$t=\frac{6.6-7.1}{\frac{1.0}{\sqrt{26}}}=-2.550$    

Step-by-step explanation:

Data given and notation  

$\bar X=6.6$ represent the sample mean

$s=1.0$ represent the sample standard deviation

$n=26$ sample size  

$\mu_o =7.1$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.  

t 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 true mean is different from 7.1 ppm, the system of hypothesis would be:  

Null hypothesis:$\mu = 7.1$  

Alternative hypothesis:$\mu \neq 7.1$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-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:  

$t=\frac{6.6-7.1}{\frac{1.0}{\sqrt{26}}}=-2.550$