In: Statistics and Probability
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.7 parts/million (ppm). A researcher believes that the current ozone level is at an insufficient level. The mean of 870 samples is 7.6 ppm. Assume a population standard deviation of 1.1. Does the data support the researcher's claim at the 0.02 level?
Step 1 of 6: State the null and alternative hypotheses.
Step 2 of 6: Find the value of the test statistic. Round your answer to two decimal places.
Step 3 of 6: Specify if the test is one-tailed or two-tailed.
Step 4 of 6: Find the P-value of the test statistic. Round your answer to four decimal places.
Step 5 of 6: Identify the level of significance for the hypothesis test.
Step 6 of 6: Make the decision to reject or fail to reject the null hypothesis.
Solution :
= 7.7
= 7.6
= 1.1
n = 870
This is the two tailed test .
The null and alternative hypothesis is ,
H0 : = 7.7
Ha : 7.7
Test statistic = z
= ( - ) / / n
= (7.6- 7.7) /1.1 / 870
= -2.68
Two tailed test
P(z < -2.68 ) = 0.0073
P-value = 0.0073
= 0.02
The critical value for a two-tailed test is zc=2.33.
p= 0.0073 < 0.02, it is concluded that the null hypothesis is rejected.
Reject the null hypothesis .
There is enough evidence to claim that the population mean μ is different than 7.7, at the 0.02 significance level.