Question

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 16 samples is 8.1 ppm with a variance of 0.64. Does the data support the claim at the 0.1 level? Assume the population distribution is approximately normal.

Step 1 of 5: State the null and alternative hypotheses.

Step 2 of 5: Find the value of the test statistic. Round your answer to three decimal places.

Step 3 of 5: Specify if the test is one-tailed or two-tailed.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: the current ozone level is not at an excess level.

Alternative hypothesis: Ha: the current ozone level is at an excess level.

H0: µ = 7.9 versus Ha: µ > 7.9

This is an upper tailed or one-tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 7.9

Xbar = 8.1

S = 0.8

n = 16

df = n – 1 = 15

α = 0.1

Critical value = 1.3406

(by using t-table or excel)

Decision rule: Reject H0 when t > 1.3406.

t = (Xbar - µ)/[S/sqrt(n)]

t = (8.1 – 7.9)/[0.8/sqrt(16)]

t = 1.0000

P-value = 0.1666

(by using t-table)

P-value > α = 0.10

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the current ozone level is at an excess level.