Question

As of January 1, 2005, the European Commission has set a limit for PM10 (particles in the air with diameters less than 10 micrometers) at 50 micrograms per cubic meter (daily average). Local health organizations in a large European city are concerned that the PM10 level in the outdoor air is higher than the 50 micrograms per cubic meter permissible. To test the validity of this statement, levels of PM10 were measured on 25 different days, yielding an average of 53.5 micrograms per cubic meter. The standard deviation for the levels of PM10 is known to be 5.8 micrograms per cubic meter. Using a z-test for the null hypothesis H0:μ=50 and the alternative hypothesis Ha:μ>50 , give the value of the test-statistic and find the p-value.

Answer 1

The z-test for the given data yields a test statistic of approximately 3.02, and the p-value is less than 0.001, which indicates strong evidence against the null hypothesis. Thus, we can conclude that the PM10 level in the outdoor air is higher than the permissible limit of 50 micrograms per cubic meter.

Step-by-step explanation:

To test the validity of the statement that the PM10 level in the outdoor air is higher than the permissible limit of 50 micrograms per cubic meter, we can conduct a z-test.

Given that the sample size is 25, the sample mean is 53.5, the population mean is 50, and the standard deviation is 5.8, we can calculate the test-statistic using the formula:

test-statistic = (sample mean - population mean) / (standard deviation / square root of sample size)

Substituting the values:

test-statistic = (53.5 - 50) / (5.8 / sqrt(25)) = 3.5 / (5.8 / 5) = 3.5 / 1.16 = 3.02 (approx).

The test-statistic is approximately 3.02. To find the p-value, we can use a z-table or a calculator. Consulting a z-table, we find that the p-value is less than 0.001, indicating strong evidence against the null hypothesis. Therefore, we can conclude that the PM10 level in the outdoor air is indeed higher than the permissible limit.