Question
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.
Hv0 : [?]
Hv1 : [?]
is this left/right or two tailed
Hv0 : [?]
Hv1 : [?]
is this left/right or two tailed
Answers
GPT-4o mini
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.
- **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.