Question

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 5.2 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 26 samples is 5.4 ppm with a variance of 0.36. Does the data support the claim at the 0.05 level? Assume the population distribution is approximately normal. Step 4 of 5: Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

Solution:
Given in the question
The claim is that the researcher believes that the current ozone level is at an excess level, so null and alternative hypothesis can be calculated as
Null hypothesis H0: = 5.2
Alternate hypothesis Ha: > 5.2
sample size(n) = 26
The sample mean (Xbar) = 5.4
Sample variance = 0.36
Sample standard deviation(S) =sqrt(Variance) = sqrt(0.36) = 0.6
Here we will use t-test because the sample size is less than 30 and the population standard deviation is unknown so the test statistic value can be calculated as
Test stat = (Xbar -)/S/sqrt(n) = (5.4-5.2)/0.6/sqrt(26) = 1.7
df = n - 1 = 26-1 = 25, and alpha=0.05, and this is right tailed test so test critical value = 1.708
Decision rule: If test stat value is greater than 1.708 than reject the null hypothesis else do not reject the null hypothesis.
As we can see that test stat value is less than 1.708, so we are failed to reject the null hypothesis.