One of the measurements used to determine the health of a person's lungs is the amount of air a person can exhale under force in one second. This is called forced expiratory volume in one second, and is abbreviated FEV1. Assume the mean FEV1 for 10-year-old boys is 2.1 liters and that the population standard deviation is o=0.4. A random sample of 78, 10 -year-old boys who live in a community with high levels of ozone pollution are found to have a sample mean FEV1 of 2.07 liters. Can you conclude that the mean FEV1 in the high-pollution community differs from 2.1 liters? Use the a=0.10 level of significance and the P-value method.

Question

Hv0 : [?]
Hv1 : [?]
is this left/right or two tailed

Answer 1

To determine whether the mean FEV1 in the high-pollution community differs from the known mean of 2.1 liters, we set up the following hypotheses:

• Null Hypothesis (H0): The mean FEV1 in the high-pollution community is equal to the population mean of 2.1 liters.

  \[ H_0: \mu = 2.1 \text{ liters} \]

• Alternative Hypothesis (H1): The mean FEV1 in the high-pollution community is not equal to 2.1 liters.

  \[ H_1: \mu \neq 2.1 \text{ liters} \]

This is a two-tailed test because we are checking for a difference in either direction (the mean could be either less than or greater than 2.1 liters).

Test Statistic Calculation

Given the known population standard deviation (\( \sigma = 0.4 \)) and the sample size (\( n = 78 \)), we can calculate the test statistic using the formula for the z-score:

\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

Where:

• \(\bar{x} = 2.07\) (sample mean)
• \(\mu = 2.1\) (hypothesized population mean)
• \(\sigma = 0.4\) (population standard deviation)
• \(n = 78\) (sample size)

Substituting in the values:

\[ z = \frac{2.07 - 2.1}{0.4 / \sqrt{78}} \]

Calculating this step-by-step:

1. Calculate \( \sigma / \sqrt{n} \):

\[ \frac{0.4}{\sqrt{78}} \approx \frac{0.4}{8.83} \approx 0.0452 \]

2. Now, compute the z-value:

\[ z = \frac{2.07 - 2.1}{0.0452} \approx \frac{-0.03}{0.0452} \approx -0.663 \]

Finding the P-value

To find the P-value for a z-score of -0.663 in a two-tailed test, we look in the standard normal distribution table (or use a calculator).

The z-score of -0.663 corresponds to a one-tailed P-value. We typically find the area to the left of -0.663 and then double it for a two-tailed test.

Using standard normal distribution tables or a calculator:

• The one-tailed P-value for z = -0.663 is approximately 0.2546.

For a two-tailed test, we double this value:

\[ \text{P-value} = 2 \times 0.2546 \approx 0.5092 \]

Conclusion

Compare the P-value to the significance level (\( \alpha = 0.10 \)):

Since \( 0.5092 > 0.10 \), we fail to reject the null hypothesis.

Final Answer

Thus, we conclude that there is not enough evidence to suggest that the mean FEV1 in the high-pollution community differs from 2.1 liters at the 0.10 level of significance.

• \( H_0: \mu = 2.1 \)
• \( H_1: \mu \neq 2.1 \)
• This is a two-tailed test.