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To determine if there is a significant difference in the population mean hours per fish caught using a boat compared to fishing from the shore at a 1% level of significance, we can conduct a paired t-test. Let's go through the process step-by-step:
### Step 1: State the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean difference between the paired observations (hours per fish caught using a boat vs. shore) is zero. Mathematically, [tex]\( \mu_D = 0 \)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean difference between the paired observations is not zero. Mathematically, [tex]\( \mu_D \neq 0 \)[/tex].
### Step 2: Data Collection and Calculation
We have the following data on hours per fish caught:
- Shore (B): 1.6, 1.8, 2.0, 3.2, 3.9, 3.6, 3.3
- Boat (A): 1.5, 1.4, 1.6, 2.2, 3.3, 3.0, 3.8
### Step 3: Perform the Paired t-test
To conduct the paired t-test, we calculate the t-statistic and the p-value to compare our sample data:
After performing these calculations, we obtain the following results:
- t-statistic: [tex]\( t = -2.083893572066533 \)[/tex]
- p-value: [tex]\( p = 0.08229037486734761 \)[/tex]
### Step 4: Compare the p-value with the Level of Significance
- Level of significance ([tex]\( \alpha \)[/tex]): 0.01
We compare the p-value to our significance level to determine if we should reject the null hypothesis.
### Step 5: Make the Decision
- If [tex]\( p \leq \alpha \)[/tex]: Reject [tex]\( H_0 \)[/tex]
- If [tex]\( p > \alpha \)[/tex]: Fail to reject [tex]\( H_0 \)[/tex]
Given that [tex]\( p = 0.08229 \)[/tex] and [tex]\( \alpha = 0.01 \)[/tex]:
Since [tex]\( p > \alpha \)[/tex], we fail to reject the null hypothesis.
### Conclusion
At the 1% level of significance, there is not enough evidence to conclude that there is a significant difference in the population mean hours per fish caught using a boat compared to fishing from the shore.
### Step 1: State the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean difference between the paired observations (hours per fish caught using a boat vs. shore) is zero. Mathematically, [tex]\( \mu_D = 0 \)[/tex].
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean difference between the paired observations is not zero. Mathematically, [tex]\( \mu_D \neq 0 \)[/tex].
### Step 2: Data Collection and Calculation
We have the following data on hours per fish caught:
- Shore (B): 1.6, 1.8, 2.0, 3.2, 3.9, 3.6, 3.3
- Boat (A): 1.5, 1.4, 1.6, 2.2, 3.3, 3.0, 3.8
### Step 3: Perform the Paired t-test
To conduct the paired t-test, we calculate the t-statistic and the p-value to compare our sample data:
After performing these calculations, we obtain the following results:
- t-statistic: [tex]\( t = -2.083893572066533 \)[/tex]
- p-value: [tex]\( p = 0.08229037486734761 \)[/tex]
### Step 4: Compare the p-value with the Level of Significance
- Level of significance ([tex]\( \alpha \)[/tex]): 0.01
We compare the p-value to our significance level to determine if we should reject the null hypothesis.
### Step 5: Make the Decision
- If [tex]\( p \leq \alpha \)[/tex]: Reject [tex]\( H_0 \)[/tex]
- If [tex]\( p > \alpha \)[/tex]: Fail to reject [tex]\( H_0 \)[/tex]
Given that [tex]\( p = 0.08229 \)[/tex] and [tex]\( \alpha = 0.01 \)[/tex]:
Since [tex]\( p > \alpha \)[/tex], we fail to reject the null hypothesis.
### Conclusion
At the 1% level of significance, there is not enough evidence to conclude that there is a significant difference in the population mean hours per fish caught using a boat compared to fishing from the shore.
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