Thank you for visiting Listed in the accompanying table are waiting times seconds of observed cars at a Delaware inspection station The data from two waiting lines are real. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To test the claim that cars in two queues (one line versus two lines) have different average waiting times, we need to perform a hypothesis test using a two-sample t-test. Here’s a step-by-step solution:
### Step 1: Formulate the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The means of the waiting times for the two populations (one line and two lines) are equal. Mathematically, [tex]\( \mu_1 = \mu_2 \)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The means of the waiting times for the two populations are not equal. Mathematically, [tex]\( \mu_1 \neq \mu_2 \)[/tex].
### Step 2: Determine the Significance Level
We are using a significance level of 0.01.
### Step 3: Calculate the Test Statistic
The test statistic for comparing two means is calculated using the formula for an independent two-sample t-test. However, the values are not computed manually in this explanation.
The computed value for the test statistic is [tex]\( t = 0.16 \)[/tex].
### Step 4: Determine the Degrees of Freedom
When using a two-sample t-test with unequal variances, we use the Welch-Satterthwaite equation to estimate the degrees of freedom. This formula isn't explicitly calculated here, but it accounts for the sample sizes and variances.
### Step 5: Calculate the p-value
The p-value is calculated using the test statistic and the degrees of freedom. For a two-tailed test, we compare the absolute value of the test statistic to a t-distribution.
The resulting p-value is [tex]\( \text{P-value} = 0.874 \)[/tex].
### Step 6: Conclusion
Compare the p-value to the significance level:
- If the p-value is less than the significance level (0.01), we reject the null hypothesis.
- If the p-value is greater than the significance level, we fail to reject the null hypothesis.
Since [tex]\( 0.874 > 0.01 \)[/tex], we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the average waiting times for the two queues are different at the 0.01 significance level.
Therefore, based on the data, we cannot claim there is a significant difference in the average waiting times between the one line and two lines at the Delaware inspection station.
### Step 1: Formulate the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The means of the waiting times for the two populations (one line and two lines) are equal. Mathematically, [tex]\( \mu_1 = \mu_2 \)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The means of the waiting times for the two populations are not equal. Mathematically, [tex]\( \mu_1 \neq \mu_2 \)[/tex].
### Step 2: Determine the Significance Level
We are using a significance level of 0.01.
### Step 3: Calculate the Test Statistic
The test statistic for comparing two means is calculated using the formula for an independent two-sample t-test. However, the values are not computed manually in this explanation.
The computed value for the test statistic is [tex]\( t = 0.16 \)[/tex].
### Step 4: Determine the Degrees of Freedom
When using a two-sample t-test with unequal variances, we use the Welch-Satterthwaite equation to estimate the degrees of freedom. This formula isn't explicitly calculated here, but it accounts for the sample sizes and variances.
### Step 5: Calculate the p-value
The p-value is calculated using the test statistic and the degrees of freedom. For a two-tailed test, we compare the absolute value of the test statistic to a t-distribution.
The resulting p-value is [tex]\( \text{P-value} = 0.874 \)[/tex].
### Step 6: Conclusion
Compare the p-value to the significance level:
- If the p-value is less than the significance level (0.01), we reject the null hypothesis.
- If the p-value is greater than the significance level, we fail to reject the null hypothesis.
Since [tex]\( 0.874 > 0.01 \)[/tex], we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the average waiting times for the two queues are different at the 0.01 significance level.
Therefore, based on the data, we cannot claim there is a significant difference in the average waiting times between the one line and two lines at the Delaware inspection station.
Thank you for reading the article Listed in the accompanying table are waiting times seconds of observed cars at a Delaware inspection station The data from two waiting lines are real. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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