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 H_{1} 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.

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