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To determine whether a simple linear regression model provides useful information for predicting growth rate from research and development expenditure, we will follow these steps:
### Part (a): Hypothesis Testing
Step 1: State the hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no significant linear relationship between growth rate and research and development expenditure ([tex]\( \beta = 0 \)[/tex]).
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): There is a significant linear relationship between growth rate and research and development expenditure ([tex]\( \beta \neq 0 \)[/tex]).
Step 2: Calculate the test statistic
From linear regression analysis, we have the slope [tex]\( \beta \)[/tex], standard error of the slope [tex]\( SE_{\beta} \)[/tex], and the t-statistic.
Given:
- Test statistic [tex]\( t = 2.28 \)[/tex]
Step 3: Determine the p-value
We use the t-statistic to find the corresponding p-value.
Given:
- [tex]\( P \text{-value} = 0.0625 \)[/tex]
Since the p-value (0.0625) is greater than the significance level of 0.05, we compare it with the significant level as follows:
Step 4: Draw a conclusion
Since the p-value (0.0625) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Conclusion:
Fail to reject [tex]\( H_0 \)[/tex]. We do not have convincing evidence of a useful linear relationship between growth rate and research and development expenditure.
### Part (b): Confidence Interval for Slope
Step 1: Calculate the confidence interval for the slope
We want to construct a 90% confidence interval for the average change in growth rate associated with a [tex]$1,000,000 increase in expenditure. Given: - slope (\( \beta \)) = 0.5751273930806419 - standard error of the slope (\( SE_{\beta} \)) = 0.25184706198095234 The critical value for a 90% confidence interval with 6 degrees of freedom (\( df = n - 2 \)) is approximately 1.943. Step 2: Compute the confidence interval The formula for the confidence interval is: \[ \text{Slope CI} = \beta \pm \text{critical t} \times SE_{\beta} \] Given: - Lower bound = 0.08576954016490956 - Upper bound = 1.0644852459963743 Hence, the confidence interval is: \[ 0.086 \leq \beta \leq 1.064\] ### Conclusion: (a) - Test Statistic: \( t = 2.28 \) - P-value: \( P = 0.0625 \) - Conclusion: Fail to reject \( H_0 \). We do not have convincing evidence of a useful linear relationship between growth rate and research and development expenditure. (b) - The 90% confidence interval for the average change in growth rate associated with a $[/tex]1,000,000 increase in expenditure is approximately [tex]\([0.086, 1.064]\)[/tex] percent per year.
### Part (a): Hypothesis Testing
Step 1: State the hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no significant linear relationship between growth rate and research and development expenditure ([tex]\( \beta = 0 \)[/tex]).
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): There is a significant linear relationship between growth rate and research and development expenditure ([tex]\( \beta \neq 0 \)[/tex]).
Step 2: Calculate the test statistic
From linear regression analysis, we have the slope [tex]\( \beta \)[/tex], standard error of the slope [tex]\( SE_{\beta} \)[/tex], and the t-statistic.
Given:
- Test statistic [tex]\( t = 2.28 \)[/tex]
Step 3: Determine the p-value
We use the t-statistic to find the corresponding p-value.
Given:
- [tex]\( P \text{-value} = 0.0625 \)[/tex]
Since the p-value (0.0625) is greater than the significance level of 0.05, we compare it with the significant level as follows:
Step 4: Draw a conclusion
Since the p-value (0.0625) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Conclusion:
Fail to reject [tex]\( H_0 \)[/tex]. We do not have convincing evidence of a useful linear relationship between growth rate and research and development expenditure.
### Part (b): Confidence Interval for Slope
Step 1: Calculate the confidence interval for the slope
We want to construct a 90% confidence interval for the average change in growth rate associated with a [tex]$1,000,000 increase in expenditure. Given: - slope (\( \beta \)) = 0.5751273930806419 - standard error of the slope (\( SE_{\beta} \)) = 0.25184706198095234 The critical value for a 90% confidence interval with 6 degrees of freedom (\( df = n - 2 \)) is approximately 1.943. Step 2: Compute the confidence interval The formula for the confidence interval is: \[ \text{Slope CI} = \beta \pm \text{critical t} \times SE_{\beta} \] Given: - Lower bound = 0.08576954016490956 - Upper bound = 1.0644852459963743 Hence, the confidence interval is: \[ 0.086 \leq \beta \leq 1.064\] ### Conclusion: (a) - Test Statistic: \( t = 2.28 \) - P-value: \( P = 0.0625 \) - Conclusion: Fail to reject \( H_0 \). We do not have convincing evidence of a useful linear relationship between growth rate and research and development expenditure. (b) - The 90% confidence interval for the average change in growth rate associated with a $[/tex]1,000,000 increase in expenditure is approximately [tex]\([0.086, 1.064]\)[/tex] percent per year.
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