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 4.5 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 1020 samples is 4.4 ppm. Assume a population standard deviation of 1.2. Does the data support the researcher's claim at the 0.05 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 place
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:
Given: The level of ozone normally found is 4.5 parts/million (ppm).
that is:
Sample Size = n = 1020
Sample mean=
Population standard deviation =
Claim: the current ozone level is not at a normal level. that is
Level of significance =
Step 1 of 6: State the null and alternative hypotheses.
Vs
Step 2 of 6: Find the value of the test statistic.
Step 3 of 6: Specify if the test is one-tailed or two-tailed.
Since H1 is not equal to type, this is two tailed test.
Step 4 of 6: Find the P-value of the test statistic.
For two tailed test, P-value is given by:
P-value = 2 X P( Z < z ) if z is negative
P-value = 2 X P( Z > z ) if z is positive
P-value = 2 X P( Z < -2.66 )
Look in z table for z = -2.6 and 0.06 and find corresponding area.
P( Z< -2.66) = 0.0039
thus
P-value = 2 X P( Z < -2.66 )
P-value = 2 X0.0039
P-value = 0.0078
Step 5 of 6: Identify the level of significance for the hypothesis test.
Level of significance =
Step 6 of 6: Make the decision to reject or fail to reject the null hypothesis.
Decision Rule:
Reject null hypothesis H0, if P-value < 0.05 level of
significance, otherwise we fail to reject H0
Since P-value = 0.0078 < 0.05 level of significance, we reject null hypothesis H0.
Thus at 0.05 level of significance, the data support the researcher's claim that the current ozone level is not at a normal level.