In: Statistics and Probability
e decimal places.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.4 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 6 samples is 7.7 ppm with a standard deviation of 0.7. Does the data support the claim at the 0.025 level? Assume the population distribution is approximately normal.
Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.4 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 6 samples is 7.7 ppm with a standard deviation of 0.7. Does the data support the claim at the 0.025 level? Assume the population distribution is approximately normal.
Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.4 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 6 samples is 7.7 ppm with a standard deviation of 0.7. Does the data support the claim at the 0.025 level? Assume the population distribution is approximately normal.
Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.4 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 6 samples is 7.7 ppm with a standard deviation of 0.7. Does the data support the claim at the 0.025 level? Assume the population distribution is approximately normal.
Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.
Answer:
Test statistic (t-value) = 1.050
Explanation:
Hypothesis
The Null hypothesis is defined as the mean ozone level is 7.4
and the Alternative Hypotheses tests the claim that the mean ozone level excess the 7.4
This is a right-tailed test.
The significance level = 0.025
Step 2 of 5
Test-statistic
Since the population is approximately normally distributed and the population standard deviation is not known, the t distribution is used to test the hypothesis,
The t statistic is obtained using the following formula,
Given:
Sample mean = 7.7
Sample standard deviation = 0.7
Sample size = 6
P-value
The p-value is obtained from t distribution table for t = 1.050 and degree of freedom = n - 1 = 6 - 1 = 5.
Conclusion
Since the P-value is great than the significance level = 0.025 at a 2.5% significance level, the null hypothesis is not rejected.