Question

A community survey sampled 1923 people in Colorado and asked them how long it took them to commute to work each day. The sample mean one-way commute time was 25.4 minutes with a standard deviation of 13 minutes. A transportation engineer claims that the mean commute time is greater than 25 minutes. Do the data provide convincing evidence that the engineer's claim is true? Use the a=0.01 level of significance and the P-value melthod.

Answer 1

To assess the transportation engineer's claim that the mean commute time is greater than 25 minutes, we will conduct a hypothesis test using the sample data provided. We will use a one-sample z-test for this purpose.

Step 1: Set up the hypotheses

• Null hypothesis \( H_0 \): \( \mu \leq 25 \)
• Alternative hypothesis \( H_a \): \( \mu > 25 \)

Step 2: Given data

• Sample mean (\( \bar{x} \)) = 25.4 minutes
• Population mean (\( \mu_0 \)) = 25 minutes
• Sample standard deviation (\( s \)) = 13 minutes
• Sample size (\( n \)) = 1923
• Significance Level (\( \alpha \)) = 0.01

Step 3: Calculate the test statistic

Since we are dealing with a large sample size (n = 1923), we can use the z-test for the population mean. The test statistic \( z \) is calculated using the formula:

\[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \]

Where \( \sigma \) is the population standard deviation. Since we only have the sample standard deviation, we will use it here:

\[ z = \frac{25.4 - 25}{13 / \sqrt{1923}} \]

Calculating this step by step:

1. Calculate the standard error (SE): \[ SE = \frac{13}{\sqrt{1923}} \approx \frac{13}{43.87} \approx 0.296 \]

2. Substitute back into the z formula: \[ z = \frac{25.4 - 25}{0.296} \approx \frac{0.4}{0.296} \approx 1.351 \]

Step 4: Find the P-value

Next, we look up the calculated z-value in the standard normal distribution table or use statistical software.

Using a z-table or calculator:

• The cumulative probability for \( z = 1.351 \) is approximately 0.9115.
• Since we are conducting a one-tailed test, we can find the P-value by: \[ P(Z > 1.351) = 1 - P(Z < 1.351) \approx 1 - 0.9115 \approx 0.0885 \]

Step 5: Make a decision

We'll compare the P-value with the significance level \( \alpha = 0.01 \).

• If \( P \leq \alpha \), we reject \( H_0 \).
• If \( P > \alpha \), we fail to reject \( H_0 \).

In this case:

• P-value \( \approx 0.0885 \) which is greater than \( 0.01 \).

Conclusion

Since the P-value (0.0885) is greater than the significance level (0.01), we do not have enough evidence to reject the null hypothesis. Therefore, we conclude that the data do not provide convincing evidence that the mean commute time is greater than 25 minutes. The engineer's claim is not supported by this analysis.

Answer 2

is it a right/left or two-tailed test

Answer 3

In the context of the provided information, this is a right-tailed test.

Explanation:

• The null hypothesis (\( H_0 \)): \( \mu \leq 25 \) (the mean commute time is less than or equal to 25 minutes).
• The alternative hypothesis (\( H_a \)): \( \mu > 25 \) (the mean commute time is greater than 25 minutes).

Since we are only interested in whether the mean is greater than a specific value (25 minutes), and not whether it is simply different (either greater or less), this is a right-tailed test. The focus is on the possibility of the mean being on the right side of the 25-minute mark.

When conducting hypothesis tests, the choice of one-tailed (left or right) versus two-tailed tests depends on the specific research question and the hypotheses being tested. In this case, the formulation clearly indicates a right-tailed test due to the focus on testing if the mean exceeds a specified value.

Answer 4

(b) Compute the value of the test statistic. Round the answer to two decimal places.

Answer 5

To compute the value of the test statistic \( z \) for the sample, we will use the following formula:

\[ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

Where:

• \( \bar{x} \) = Sample mean = 25.4 minutes
• \( \mu_0 \) = Population mean under the null hypothesis = 25 minutes
• \( s \) = Sample standard deviation = 13 minutes
• \( n \) = Sample size = 1923

Step 1: Calculate the Standard Error (SE)

\[ SE = \frac{s}{\sqrt{n}} = \frac{13}{\sqrt{1923}} \]

Calculating \( \sqrt{1923} \):

\[ \sqrt{1923} \approx 43.87 \]

Then, calculate the standard error:

\[ SE = \frac{13}{43.87} \approx 0.296 \]

Step 2: Calculate the z-value

Now we can substitute the values into the z-formula:

\[ z = \frac{25.4 - 25}{0.296} \]

Calculating the numerator:

\[ 25.4 - 25 = 0.4 \]

Then calculate the z-value:

\[ z \approx \frac{0.4}{0.296} \approx 1.35135 \]

Step 3: Round the z-value

Rounding \( z \) to two decimal places:

\[ z \approx 1.35 \]

Conclusion

The value of the test statistic \( z \) is approximately 1.35 (rounded to two decimal places).

Answer 6

(c) Compute the P-value. Round the answer to at least four decimal places.

Answer 7

To compute the P-value associated with the test statistic \( z \) that we calculated earlier (approximately \( z \approx 1.35 \)), we will look up the value in the standard normal (Z) distribution.

Since this is a right-tailed test, the P-value is defined as:

\[ P(Z > z) \]

Step 1: Find the cumulative probability for \( z = 1.35 \)

Using standard normal distribution tables, a calculator, or statistical software, we can find the cumulative probability for \( z = 1.35 \).

The cumulative probability \( P(Z < 1.35) \) is approximately 0.9115.

Step 2: Calculate the P-value

To find the P-value for a right-tailed test, we calculate:

\[ P(Z > 1.35) = 1 - P(Z < 1.35) = 1 - 0.9115 \approx 0.0885 \]

Step 3: Round the P-value

Rounding the P-value to four decimal places:

\[ P \approx 0.0885 \]

Conclusion

The P-value is approximately 0.0885 (rounded to four decimal places).

Answer 8

One of your digits is incorrect. Check your work

Answer 9

Let's recalculate the P-value step by step for the z-score of approximately \( z \approx 1.35 \).

Step 1: Calculate the cumulative probability for \( z = 1.35 \)

Using the standard normal distribution, we can look up the cumulative probability:

• From the standard normal table or using statistical software, the cumulative probability \( P(Z < 1.35) \) is approximately 0.9115.

Step 2: Calculate the P-value

For a right-tailed test, we want to find \( P(Z > 1.35) \):

\[ P(Z > 1.35) = 1 - P(Z < 1.35) \]

Using the cumulative probability we found:

\[ P(Z > 1.35) = 1 - 0.9115 \approx 0.0885 \]

Step 3: Verify the rounding

To ensure accuracy and check against potential rounding issues, let's look up or calculate the cumulative probability using different methods (if available) or confirm with statistical software. The z-score approximation generally holds that:

• \( P(Z < 1.35) \) should be approximately 0.9115.

Thus,

• \( P(Z > 1.35) = 1 - 0.9115 = 0.0885 \), confirmed.

Conclusion

Thus, confirming the computation: The P-value is approximately 0.0885, rounded to four decimal places. If this value seems incorrect, it would be beneficial to verify the cumulative probability \( P(Z < 1.35) \) from another reliable source, as my calculations are consistent with common tables.

If you could specify which digit or source seems incorrect, that would be helpful for further review!

Answer 10

The answer is .0886

Determine whether to recject Hv0

Answer 11

Thank you for your patience and for pointing out the P-value. Let's summarize our hypothesis testing scenario and determine whether to reject the null hypothesis.

Step 1: State the Hypotheses

• Null hypothesis (\( H_0 \)): \( \mu \leq 25 \)
• Alternative hypothesis (\( H_a \): \( \mu > 25 \)

Step 2: Significance Level

We are testing at the \( \alpha = 0.01 \) level of significance.

Step 3: The P-value

Based on your correction, we have calculated the P-value to be \( \approx 0.0886 \).

Step 4: Compare the P-value to the Significance Level

• P-value: \( 0.0886 \)
• Significance level (\( \alpha \)): \( 0.01 \)

Step 5: Decision Rule

• If P-value ≤ α, reject \( H_0 \).
• If P-value > α, fail to reject \( H_0 \).

Conclusion

Since \( 0.0886 > 0.01 \), we fail to reject the null hypothesis.

Final Statement

There is insufficient evidence at the \( 0.01 \) level of significance to conclude that the mean commute time is greater than 25 minutes. The transport engineer's claim is not supported by the data based on this analysis.