Problem 35
Question
A regional commuter airline selected a random sample of 25 flights and found that the correlation between the number of passengers and the total weight, in pounds, of luggage stored in the luggage compartment is 0.94 Using the .05 significance level, can we conclude that there is a positive association between the two variables?
Step-by-Step Solution
Verified Answer
Yes, we can conclude there is a positive association.
1Step 1: State Hypotheses
First, we need to set up our hypotheses. The null hypothesis (H0) is that there is no correlation between the number of passengers and the total luggage weight, i.e., \( \rho = 0 \). The alternative hypothesis (H1) is that there is a positive correlation, i.e., \( \rho > 0 \).
2Step 2: Determine Critical Value
We need to determine the critical value for the correlation coefficient, given the significance level and degrees of freedom. Since we have a sample of 25 flights, the degrees of freedom is \( n - 2 = 23 \). For a one-tailed test at \( \alpha = 0.05 \), check a t-table or a correlation critical value table to find the critical value for \( r \).
3Step 3: Calculate Test Statistic
Using the formula for the test statistic for correlation \( t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \), where \( r = 0.94 \) and \( n = 25 \), calculate the test statistic. Substituting the values yields \( t = \frac{0.94 \sqrt{23}}{\sqrt{1-(0.94)^2}} \). Compute this value to find \( t \).
4Step 4: Compare Test Statistic to Critical Value
Determine if the computed test statistic exceeds the critical value found in Step 2. If the test statistic is greater than the critical value, we can reject the null hypothesis.
5Step 5: Make a Conclusion
Based on the comparison in Step 4, conclude whether the correlation between the number of passengers and the total luggage weight is statistically significant at the 0.05 level. If you reject the null hypothesis, it means there is significant evidence of a positive association between the variables.
Key Concepts
Hypothesis TestingSignificance LevelCritical ValueTest Statistic
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on a sample. At its core, it involves comparing two competing hypotheses:
This process provides a structured way of making inferences about a population, ensuring our conclusions are backed by data.
- The null hypothesis ( \( H_0 \)), which typically states that there is no effect or no difference. In our exercise, it is that there is no correlation between the two variables, meaning \( \rho = 0 \).
- The alternative hypothesis ( \( H_1 \)), suggesting that there is an effect or a difference. Here, it suggests a positive correlation, or \( \rho > 0 \).
This process provides a structured way of making inferences about a population, ensuring our conclusions are backed by data.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold set by the researcher to decide whether to reject the null hypothesis.
This value represents the probability of committing a Type I error, which means rejecting a true null hypothesis. A common choice for \( \alpha \) is 0.05, as used in our exercise, implying a 5% risk of making such an error.
When planning an experiment, \( \alpha \) is selected before analyzing the data, providing a clear benchmark for what constitutes significant evidence against \( H_0 \).
In essence, it helps us control how risk-averse we want to be in drawing conclusions. A smaller \( \alpha \) decreases the chance of error but requires stronger evidence to reject \( H_0 \).
This value represents the probability of committing a Type I error, which means rejecting a true null hypothesis. A common choice for \( \alpha \) is 0.05, as used in our exercise, implying a 5% risk of making such an error.
When planning an experiment, \( \alpha \) is selected before analyzing the data, providing a clear benchmark for what constitutes significant evidence against \( H_0 \).
In essence, it helps us control how risk-averse we want to be in drawing conclusions. A smaller \( \alpha \) decreases the chance of error but requires stronger evidence to reject \( H_0 \).
Critical Value
A critical value helps determine the threshold at which we can reject the null hypothesis.
It is based on the chosen significance level \( \alpha \) and the sample's degrees of freedom ( \( df \)). For our example, with 25 flights, we have \( df = n - 2 = 23 \).
The critical value for correlation testing can be found in statistical tables or using software tools. It represents the cutoff point for the test statistic. If the calculated test statistic exceeds this critical value, \( H_0 \) is rejected.
This pivotal number thus acts as a gatekeeper, helping us decide whether the apparent association observed (correlation coefficient here) is strong enough to be deemed statistically significant.
It is based on the chosen significance level \( \alpha \) and the sample's degrees of freedom ( \( df \)). For our example, with 25 flights, we have \( df = n - 2 = 23 \).
The critical value for correlation testing can be found in statistical tables or using software tools. It represents the cutoff point for the test statistic. If the calculated test statistic exceeds this critical value, \( H_0 \) is rejected.
This pivotal number thus acts as a gatekeeper, helping us decide whether the apparent association observed (correlation coefficient here) is strong enough to be deemed statistically significant.
Test Statistic
A test statistic is a standardized value that is derived from sample data during a hypothesis test.
For correlation, the test statistic is computed using the formula:\[t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}}\]where \( r \) is the sample correlation coefficient, and \( n \) is the number of observations. In our problem, \( r = 0.94 \) and \( n = 25 \).
Substituting these values into the formula helps determine whether the observed correlation in the sample is strong enough to reject the null hypothesis.
The value obtained here is then compared to the critical value. If it is larger, the correlation observed is statistically significant, supporting \( H_1 \) further.
For correlation, the test statistic is computed using the formula:\[t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}}\]where \( r \) is the sample correlation coefficient, and \( n \) is the number of observations. In our problem, \( r = 0.94 \) and \( n = 25 \).
Substituting these values into the formula helps determine whether the observed correlation in the sample is strong enough to reject the null hypothesis.
The value obtained here is then compared to the critical value. If it is larger, the correlation observed is statistically significant, supporting \( H_1 \) further.
Other exercises in this chapter
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