| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if the results conform to the expectation, we can use a one-sample t-test. The null hypothesis (H0) is that the mean yield of the new variety green gram is equal to the expected yield of 12.0 quintals per hectare. The alternative hypothesis (H1) is that the mean yield is not equal to 12.0 quintals per hectare. First, let's calculate the mean yield of the 10 farmers' fields: Mean = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 = 126.3 / 10 = 12.63 quintals per hectare Next, we calculate the standard deviation of the yields: Standard deviation = sqrt(((14.3-12.63)^2 + (12.6-12.63)^2 + (13.7-12.63)^2 + (10.9-12.63)^2 + (13.7-12.63)^2 + (12.0-12.63)^2 + (11.4-12.63)^2 + (12.0-12.63)^2 + (12.6-12.63)^2 + (13.1-12.63)^2) / 9) = sqrt(3.466666667) = 1.86185 Now we calculate the t-statistic: t = (mean - expected yield) / (standard deviation / sqrt(sample size)) = (12.63 - 12.0) / (1.86185 / sqrt(10)) = 1.236 Using a t-table or calculator, we find the critical t-value for a two-tailed test with 9 degrees of freedom (sample size - 1) and a 5% level of significance to be approximately ±2.262. Since the calculated t-statistic (1.236) falls within the range of the critical t-values (-2.262 to 2.262), we fail to reject the null hypothesis. This means that the results do not provide enough evidence to conclude that the mean yield of the new variety green gram is significantly different from the expected yield of 12.0 quintals per hectare at the 5% level of significance. Therefore, the results conform to the expectation. |
True |
| deepseekr1 | To determine if the sample data conforms to the expected yield of 12.0 quintals/hectare at a 5% significance level, we perform a one-sample t-test. **Step 1: Hypotheses** - **Null Hypothesis (H₀):** μ = 12.0 (The mean yield is 12.0 quintals/hectare) - **Alternative Hypothesis (H₁):** μ ≠ 12.0 (The mean yield differs from 12.0 quintals/hectare) **Step 2: Calculate Sample Mean and Standard Deviation** - **Sample Mean (x̄):** \[ x̄ = \frac{\sum x_i}{n} = \frac{14.3 + 12.6 + \dots + 13.1}{10} = 12.63 \, \text{quintals/hectare} \] - **Sample Standard Deviation (s):** \[ s = \sqrt{\frac{\sum (x_i - x̄)^2}{n-1}} = \sqrt{\frac{10.601}{9}} \approx 1.085 \, \text{quintals/hectare} \] **Step 3: Compute t-Statistic** \[ t = \frac{x̄ - μ}{s / \sqrt{n}} = \frac{12.63 - 12.0}{1.085 / \sqrt{10}} \approx \frac{0.63}{0.343} \approx 1.836 \] **Step 4: Determine Critical Value and Compare** - **Degrees of Freedom (df):** \( n - 1 = 9 \) - **Critical t-value (α = 0.05, two-tailed):** \( ±2.262 \) - **Conclusion:** Since \( |t| = 1.836 < 2.262 \), we **fail to reject H₀**. **Step 5: Interpretation** The sample data does not provide sufficient evidence to conclude that the mean yield differs from 12.0 quintals/hectare at the 5% significance level. **Final Answer:** \boxed{\text{The results conform to the expectation at the 5\% significance level.}} |
True |
| gpto1 | **Solution:** To determine whether the observed yields conform to the expected yield of 12.0 quintals per hectare at a 5% significance level, we will perform a hypothesis test using the sample data from the 10 farmers' fields. **Step 1: State the Hypotheses** - **Null Hypothesis (H₀):** The true mean yield of the new green gram variety is 12.0 quintals per hectare. \( H_0: \mu = 12.0 \) - **Alternative Hypothesis (H₁):** The true mean yield is not equal to 12.0 quintals per hectare. \( H_1: \mu \ne 12.0 \) This is a two-tailed test since we are checking for any significant difference (either higher or lower) from the expected mean. **Step 2: Collect and Summarize the Sample Data** The yields (in quintals per hectare) from the 10 farmers' fields are: 1. 14.3 2. 12.6 3. 13.7 4. 10.9 5. 13.7 6. 12.0 7. 11.4 8. 12.0 9. 12.6 10. 13.1 **Calculate the Sample Mean (\( \bar{x} \))** \( \bar{x} = \frac{\text{Sum of all sample yields}}{n} \) \( \bar{x} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} \) \( \bar{x} = \frac{126.3}{10} \) \( \bar{x} = 12.63 \) quintals per hectare **Calculate the Sample Standard Deviation (s)** First, calculate each deviation from the mean and square it. | \( x_i \) | \( (x_i - \bar{x}) \) | \( (x_i - \bar{x})^2 \) | |-----------|------------------------|-------------------------| | 14.3 | 14.3 - 12.63 = 1.67 | \( (1.67)^2 = 2.7889 \) | | 12.6 | 12.6 - 12.63 = -0.03 | \( (-0.03)^2 = 0.0009 \) | | 13.7 | 13.7 - 12.63 = 1.07 | \( (1.07)^2 = 1.1449 \) | | 10.9 | 10.9 - 12.63 = -1.73 | \( (-1.73)^2 = 2.9929 \) | | 13.7 | 13.7 - 12.63 = 1.07 | \( (1.07)^2 = 1.1449 \) | | 12.0 | 12.0 - 12.63 = -0.63 | \( (-0.63)^2 = 0.3969 \) | | 11.4 | 11.4 - 12.63 = -1.23 | \( (-1.23)^2 = 1.5129 \) | | 12.0 | 12.0 - 12.63 = -0.63 | \( (-0.63)^2 = 0.3969 \) | | 12.6 | 12.6 - 12.63 = -0.03 | \( (-0.03)^2 = 0.0009 \) | | 13.1 | 13.1 - 12.63 = 0.47 | \( (0.47)^2 = 0.2209 \) | **Sum of Squared Deviations (SSD):** \( \text{SSD} = 2.7889 + 0.0009 + 1.1449 + 2.9929 + 1.1449 + 0.3969 + 1.5129 + 0.3969 + 0.0009 + 0.2209 \) \( \text{SSD} = 10.6012 \) (rounded to four decimal places) **Calculate the Sample Variance (s²):** \( s^2 = \frac{\text{SSD}}{n - 1} = \frac{10.6012}{10 - 1} = \frac{10.6012}{9} = 1.1780 \) **Calculate the Sample Standard Deviation (s):** \( s = \sqrt{s^2} = \sqrt{1.1780} = 1.0854 \) quintals per hectare **Step 3: Calculate the Test Statistic** Since the population standard deviation is unknown and the sample size is small (\( n < 30 \)), we will use the t-test. **Calculate the Standard Error (SE):** \( \text{SE} = \frac{s}{\sqrt{n}} = \frac{1.0854}{\sqrt{10}} = \frac{1.0854}{3.1623} = 0.3435 \) **Calculate the t-statistic:** \( t = \frac{\bar{x} - \mu_0}{\text{SE}} = \frac{12.63 - 12.0}{0.3435} = \frac{0.63}{0.3435} = 1.834 \) **Step 4: Determine the Critical t-value** Degrees of freedom (df): \( \text{df} = n - 1 = 10 - 1 = 9 \) For a two-tailed test at a 5% significance level (\( \alpha = 0.05 \)), we look up the critical t-value from the t-distribution table for \( df = 9 \). \( t_{\text{critical}} = t_{(\alpha/2, \, df)} = t_{(0.025, \,9)} \) From the t-distribution table: \( t_{\text{critical}} = \pm 2.262 \) **Step 5: Make a Decision** - **If \( |t_{\text{calculated}}| > t_{\text{critical}} \)**, we reject the null hypothesis. - **If \( |t_{\text{calculated}}| \leq t_{\text{critical}} \)**, we fail to reject the null hypothesis. Calculated t-statistic: \( |t_{\text{calculated}}| = |1.834| = 1.834 \) Since: \( 1.834 < 2.262 \) We **fail to reject the null hypothesis**. **Step 6: Conclusion** At the 5% significance level, there is no sufficient evidence to conclude that the true mean yield of the new green gram variety is different from 12.0 quintals per hectare. Therefore, the observed yields **conform to the expectation**. **Answer:** Yes, the results conform to the expectation at the 5% significance level. The statistical analysis shows that there is no significant difference between the observed mean yield and the expected yield of 12.0 quintals per hectare. |
True |
| GPT-4o | To determine whether the observed yield data conform to the expected yield of 12.0 quintals per hectare, we can perform a hypothesis test. Specifically, we will use a one-sample t-test to compare the sample mean with the population mean (expected yield). Here are the steps to follow: ### Step 1: State the Hypotheses - Null Hypothesis (\(H_0\)): The mean yield of the new variety of green gram is 12.0 quintals per hectare. \[ H_0: \mu = 12.0 \] - Alternative Hypothesis (\(H_1\)): The mean yield of the new variety of green gram is not 12.0 quintals per hectare. \[ H_1: \mu \neq 12.0 \] ### Step 2: Collect the Sample Data The yields from the 10 randomly selected farmers' fields are: \[ 14.3, 12.6, 13.7, 10.9, 13.7, 12.0, 11.4, 12.0, 12.6, 13.1 \] ### Step 3: Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (\(s\)) The sample mean (\(\bar{x}\)) is calculated as follows: \[ \bar{x} = \frac{\sum x_i}{n} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} = \frac{126.3}{10} = 12.63 \] The sample standard deviation (\(s\)) is calculated using the formula: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \] First, find the squared differences: \[ (14.3 - 12.63)^2, (12.6 - 12.63)^2, \ldots, (13.1 - 12.63)^2 \] \[ = 2.7889, 0.0009, 1.1449, 2.9857, 1.1449, 0.3969, 1.4977, 0.3969, 0.0009, 0.2197 \] Sum of squared differences: \[ \sum (x_i - \bar{x})^2 = 10.5775 \] Now, calculate the sample standard deviation: \[ s = \sqrt{\frac{10.5775}{10 - 1}} = \sqrt{1.1753} \approx 1.0842 \] ### Step 4: Calculate the Test Statistic (t) The test statistic for a one-sample t-test is given by: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] Substitute the known values: \[ t = \frac{12.63 - 12.0}{1.0842 / \sqrt{10}} = \frac{0.63}{0.3429} \approx 1.84 \] ### Step 5: Determine the Degrees of Freedom (df) and Critical t-Value Degrees of freedom (\(df\)) is: \[ df = n - 1 = 10 - 1 = 9 \] For a two-tailed test at the 5% significance level (\(\alpha = 0.05\)), the critical t-value can be found using a t-distribution table or calculator. For \(df = 9\): \[ t_{critical} \approx \pm 2.262 \] ### Step 6: Compare the Test Statistic with the Critical t-Value \[ |t| = 1.84 \] \[ t_{critical} = 2.262 \] Since \( |t| < t_{critical} \), we fail to reject the null hypothesis. ### Step 7: Conclusion At the 5% significance level, there is not enough evidence to reject the null hypothesis. Therefore, the results conform to the expectation that the mean yield of the new variety of green gram is 12.0 quintals per hectare. |
True |
| llama3.1-70B-Instruct | To determine if the results conform to the expectation, we can perform a one-sample t-test. The null hypothesis (H0) is that the mean yield of the new variety green gram is equal to the expected yield of 12.0 quintals per hectare, while the alternative hypothesis (H1) is that the mean yield is not equal to 12.0 quintals per hectare. Here are the steps to perform the t-test: 1. Calculate the sample mean (x̄) and sample standard deviation (s) of the yield data: x̄ = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 x̄ = 12.53 s = √[(Σ(xi - x̄)^2) / (n - 1)] s = √[(14.3 - 12.53)^2 + (12.6 - 12.53)^2 + ... + (13.1 - 12.53)^2] / 9 s = 1.13 2. Calculate the t-statistic: t = (x̄ - μ) / (s / √n) t = (12.53 - 12.0) / (1.13 / √10) t = 0.53 / 0.36 t = 1.47 3. Determine the degrees of freedom (df) and the critical t-value for the given level of significance (α = 0.05): df = n - 1 = 10 - 1 = 9 Using a t-distribution table or calculator, we find that the critical t-value for α = 0.05 and df = 9 is approximately 2.262. 4. Compare the calculated t-statistic to the critical t-value: Since |t| = 1.47 is less than the critical t-value of 2.262, we fail to reject the null hypothesis (H0). Conclusion: Based on the one-sample t-test, we conclude that the results conform to the expectation of a mean yield of 12.0 quintals per hectare at a 5% level of significance. The observed mean yield of 12.53 quintals per hectare is not significantly different from the expected yield. |
True |
| llama3.1-405B-Instruct | To determine if the results conform to the expected yield of 12.0 quintals per hectare, we can perform a one-sample t-test. Here's the step-by-step solution: 1. State the null and alternative hypotheses: H0: μ = 12.0 (The true mean yield is equal to 12.0 quintals per hectare) H1: μ ≠ 12.0 (The true mean yield is not equal to 12.0 quintals per hectare) 2. Calculate the sample mean (x̄) and sample standard deviation (s): x̄ = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 = 12.53 s = √[(Σ(xi - x̄)^2) / (n - 1)] = √[(Σ(xi - 12.53)^2) / 9] = √[(14.3 - 12.53)^2 + (12.6 - 12.53)^2 + ... + (13.1 - 12.53)^2) / 9] = 1.23 3. Calculate the standard error (SE): SE = s / √n = 1.23 / √10 = 0.39 4. Calculate the t-statistic: t = (x̄ - μ) / SE = (12.53 - 12.0) / 0.39 = 1.35 5. Determine the degrees of freedom (df): df = n - 1 = 10 - 1 = 9 6. Look up the critical t-value in a t-distribution table or use a calculator/software to find the p-value: Using a t-distribution table, we find that the critical t-value for a two-tailed test with df = 9 and α = 0.05 is approximately 2.262. Using a calculator/software, we find that the p-value associated with the t-statistic (1.35) and df = 9 is approximately 0.21. 7. Compare the calculated t-statistic to the critical t-value or p-value: Since the calculated t-statistic (1.35) is less than the critical t-value (2.262), we fail to reject the null hypothesis. Alternatively, since the p-value (0.21) is greater than the significance level (0.05), we fail to reject the null hypothesis. Conclusion: Based on the sample data, we fail to reject the null hypothesis that the true mean yield is equal to 12.0 quintals per hectare. This suggests that the results conform to the expected yield of 12.0 quintals per hectare at a 5% level of significance. |
True |
| DeepSeek-Coder-V2 | To determine if the results conform to the expectation of a yield of 12.0 quintals per hectare with a level of significance of 5%, we can perform a one-sample t-test. Here are the steps to solve the problem: 1. **State the Hypotheses:** - Null Hypothesis (H0): The mean yield is equal to 12.0 quintals per hectare. \[ H_0: \mu = 12.0 \] - Alternative Hypothesis (H1): The mean yield is not equal to 12.0 quintals per hectare. \[ H_1: \mu \neq 12.0 \] 2. **Calculate the Sample Mean (\(\bar{X}\)):** The yields recorded are: 14.3, 12.6, 13.7, 10.9, 13.7, 12.0, 11.4, 12.0, 12.6, 13.1. \[ \bar{X} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} \] \[ \bar{X} = \frac{126.3}{10} = 12.63 \] 3. **Calculate the Sample Standard Deviation (s):** First, calculate the sum of squares (SS): \[ SS = \sum (X_i - \bar{X})^2 \] \[ SS = (14.3 - 12.63)^2 + (12.6 - 12.63)^2 + (13.7 - 12.63)^2 + (10.9 - 12.63)^2 + (13.7 - 12.63)^2 + (12.0 - 12.63)^2 + (11.4 - 12.63)^2 + (12.0 - 12.63)^2 + (12.6 - 12.63)^2 + (13.1 - 12.63)^2 \] \[ SS = 2.8089 + 0.0009 + 1.1449 + 2.9929 + 1.1449 + 0.3969 + 1.5129 + 0.3969 + 0.0009 + 0.2209 \] \[ SS = 10.621 \] \[ s = \sqrt{\frac{SS}{n-1}} = \sqrt{\frac{10.621}{9}} = \sqrt{1.1801} \approx 1.086 \] 4. **Calculate the t-statistic:** \[ t = \frac{\bar{X} - \mu}{s / \sqrt{n}} \] \[ t = \frac{12.63 - 12.0}{1.086 / \sqrt{10}} \] \[ t = \frac{0.63}{1.086 / 3.162} = \frac{0.63}{0.343} \approx 1.837 \] 5. **Determine the Critical t-value:** For a two-tailed test with 9 degrees of freedom (df = n - 1 = 10 - 1 = 9) and a significance level of 5%, the critical t-values are approximately ±2.262. 6. **Compare the t-statistic with the critical t-value:** \[ |t| = 1.837 < 2.262 \] Since the calculated t-statistic (1.837) is less than the critical t-value (2.262), we fail to reject the null hypothesis. **Conclusion:** At the 5% level of significance, there is not enough evidence to conclude that the mean yield is different from 12.0 quintals per hectare. The results conform to the expectation. |
True |
| Qwen2-72B-Instruct | To determine if the results conform to the expectation, we can use a one-sample t-test. The null hypothesis (H0) is that the mean yield of the new variety green gram is equal to the expected yield of 12.0 quintals per hectare. The alternative hypothesis (H1) is that the mean yield is not equal to 12.0 quintals per hectare. First, let's calculate the mean yield of the 10 farmers' fields: Mean = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 = 126.3 / 10 = 12.63 quintals per hectare Next, we calculate the standard deviation of the yields: Standard deviation = sqrt(((14.3-12.63)^2 + (12.6-12.63)^2 + (13.7-12.63)^2 + (10.9-12.63)^2 + (13.7-12.63)^2 + (12.0-12.63)^2 + (11.4-12.63)^2 + (12.0-12.63)^2 + (12.6-12.63)^2 + (13.1-12.63)^2) / 9) = sqrt(3.466666667) = 1.86185 Now we calculate the t-statistic: t = (mean - expected yield) / (standard deviation / sqrt(sample size)) = (12.63 - 12.0) / (1.86185 / sqrt(10)) = 1.236 Using a t-table or calculator, we find the critical t-value for a two-tailed test with 9 degrees of freedom (sample size - 1) and a 5% level of significance to be approximately ±2.262. Since the calculated t-statistic (1.236) falls within the range of the critical t-values (-2.262 to 2.262), we fail to reject the null hypothesis. This means that the results do not provide enough evidence to conclude that the mean yield of the new variety green gram is significantly different from the expected yield of 12.0 quintals per hectare at the 5% level of significance. Therefore, the results conform to the expectation. |
True |
| baichuan4 | To determine if the results conform to the expectation with a 5% level of significance, we can use a one-sample t-test to compare the mean yield of the new variety of green gram to the expected yield of 12.0 quintals per hectare. First, let's calculate the sample mean (\(\bar{x}\)) and sample standard deviation (s) of the observed yields: Sample Mean (\(\bar{x}\)): \[ \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} \] \[ \bar{x} = 12.61 \] Sample Standard Deviation (s): \[ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2} \] \[ s = \sqrt{\frac{(14.3 - 12.61)^2 + (12.6 - 12.61)^2 + \cdots + (13.1 - 12.61)^2}{10-1}} \] \[ s = \sqrt{\frac{3.7962}{9}} \] \[ s = 0.668 \] Now, we can calculate the t-statistic: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] where \(\mu_0\) is the expected yield (12.0 quintals per hectare), \(\bar{x}\) is the sample mean (12.61 quintals per hectare), s is the sample standard deviation (0.668 quintals per hectare), and n is the sample size (10). \[ t = \frac{12.61 - 12.0}{0.668/\sqrt{10}} \] \[ t = \frac{0.61}{0.212} \] \[ t = 2.88 \] With 9 degrees of freedom (n-1), we can look up the critical value for a one-tailed t-test at the 5% significance level. If our calculated t-statistic exceeds this critical value, we can reject the null hypothesis that the true mean yield is equal to 12.0 quintals per hectare. From a t-table or using a statistical software, we find that the critical value for a one-tailed t-test with 9 degrees of freedom at the 5% significance level is approximately 1.833. Since our calculated t-statistic (2.88) exceeds this critical value, we can reject the null hypothesis and conclude that the results do not conform to the expectation of 12.0 quintals per hectare. Instead, the data suggests that the new variety of green gram has a significantly higher yield than expected. |
True |
| doubao | First, we calculate the mean of the observed yields: Mean = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 = 12.56 quintals/hectare. Next, we calculate the standard deviation of the observed yields. Then we use a one-sample t-test to compare the mean yield with the expected yield of 12.0 quintals/hectare. The critical value for a one-sample t-test with 9 degrees of freedom (10 - 1) at a 5% significance level is approximately ±2.262. The t-value is calculated as (mean - expected yield) / (standard deviation / √10). If the absolute value of the t-value is less than the critical value, we can conclude that the results conform to the expectation. Otherwise, we reject the hypothesis. After performing the calculations, if the t-value is within the acceptable range, then the results conform to the expectation; otherwise, they do not. |
True |
| minimax | To determine whether the results conform to the expectation, we can perform a one-sample t-test. The null hypothesis (H0) is that the true mean yield of the new variety of green gram is equal to the expected yield of 12.0 quintals per hectare. The alternative hypothesis (H1) is that the true mean yield is not equal to 12.0 quintals per hectare. Given data: - Expected mean yield (μ0) = 12.0 quintals/hectare - Sample mean (x̄) = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 - Sample standard deviation (s) = sqrt(Σ(xi - x̄)² / (n - 1)) - Number of observations (n) = 10 - Level of significance (α) = 0.05 First, we calculate the sample mean: x̄ = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 x̄ = 121.3 / 10 x̄ = 12.13 quintals/hectare Next, we calculate the sample standard deviation: s = sqrt(Σ(xi - x̄)² / (n - 1)) s = sqrt(((14.3 - 12.13)² + (12.6 - 12.13)² + ... + (13.1 - 12.13)²) / (10 - 1)) Let's calculate the sum of squared differences: Σ(xi - x̄)² = (14.3 - 12.13)² + (12.6 - 12.13)² + ... + (13.1 - 12.13)² = (2.17)² + (0.47)² + (1.57)² + (-1.23)² + (1.57)² + (-0.13)² + (-0.73)² + (-0.13)² + (0.47)² + (0.97)² = 4.7089 + 0.2209 + 2.4649 + 1.5129 + 2.4649 + 0.0169 + 0.5329 + 0.0169 + 0.2209 + 0.9409 = 13.0961 Now, we can calculate the sample standard deviation: s = sqrt(13.0961 / (10 - 1)) s = sqrt(13.0961 / 9) s = sqrt(1.4551) s ≈ 1.2063 quintals/hectare Now we can calculate the t-statistic: t = (x̄ - μ0) / (s / sqrt(n)) t = (12.13 - 12.0) / (1.2063 / sqrt(10)) t = 0.13 / (1.2063 / 3.1623) t ≈ 0.13 / 0.3814 t ≈ 0.3409 The degrees of freedom (df) for this test is n - 1, which is 10 - 1 = 9. Now we need to compare the calculated t-statistic to the critical t-value from the t-distribution table at a 5% level of significance for a two-tailed test with 9 degrees of freedom. Looking up the critical t-value for df = 9 and α = 0.05 (two-tailed), we find that the critical t-value is approximately 2.262. Since the calculated t-statistic (0.3409) is less than the critical t-value (2.262), we fail to reject the null hypothesis. This means that the results do conform to the expectation that the true mean yield of the new variety of green gram is 12.0 quintals per hectare at the 5% level of significance. |
True |
| kimi | To determine if the results conform to the expectation with a level of significance of 5%, we can perform a one-sample t-test. The null hypothesis (H0) is that the mean yield is equal to the expected yield, and the alternative hypothesis (H1) is that the mean yield is not equal to the expected yield. Here are the steps to perform the test: 1. **State the Hypotheses:** - H0: μ = μ0 = 12.0 (The mean yield is equal to the expected yield) - H1: μ ≠ μ0 (The mean yield is not equal to the expected yield) 2. **Calculate the Sample Mean (x̄):** \[ \bar{x} = \frac{\sum{x_i}}{n} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} = \frac{122.3}{10} = 12.23 \] 3. **Calculate the Sample Standard Deviation (s):** \[ s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \] \[ s = \sqrt{\frac{(14.3 - 12.23)^2 + (12.6 - 12.23)^2 + (13.7 - 12.23)^2 + (10.9 - 12.23)^2 + (13.7 - 12.23)^2 + (12.0 - 12.23)^2 + (11.4 - 12.23)^2 + (12.0 - 12.23)^2 + (12.6 - 12.23)^2 + (13.1 - 12.23)^2}{9}} \] \[ s = \sqrt{\frac{(2.07)^2 + (0.37)^2 + (1.47)^2 + (-1.33)^2 + (1.47)^2 + (-0.23)^2 + (-0.83)^2 + (-0.23)^2 + (0.37)^2 + (0.87)^2}{9}} \] \[ s = \sqrt{\frac{4.2849 + 0.1369 + 2.1609 + 1.7689 + 2.1609 + 0.0529 + 0.6889 + 0.0529 + 0.1369 + 0.7569}{9}} \] \[ s = \sqrt{\frac{12.2849}{9}} = \sqrt{1.3649} \approx 1.167 \] 4. **Calculate the t-statistic:** \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{12.23 - 12.0}{1.167 / \sqrt{10}} = \frac{0.23}{0.37} \approx 0.621 \] 5. **Determine the Critical t-value:** For a two-tailed test with a 5% level of significance and 9 degrees of freedom (n-1 = 10-1 = 9), the critical t-value can be found using t-distribution tables or a calculator. The critical t-value is approximately ±2.262. 6. **Compare the t-statistic to the critical t-value:** Since the calculated t-statistic (0.621) is less than the critical t-value (±2.262), we fail to reject the null hypothesis. 7. **Conclusion:** At a 5% level of significance, there is not enough evidence to reject the null hypothesis. Therefore, the results conform to the expectation. This means that based on the data collected, the new variety of green gram does not significantly differ from the expected yield of 12.0 quintals per hectare. |
True |
| wenxinyiyan | To solve this problem, we will use a One-Sample T-Test to determine if the average yield of the new variety green gram conforms to the expected yield of 12.0 quintals per hectare, with a significance level of 5%. The null hypothesis (H0) is that the average yield is equal to 12.0 quintals per hectare, and the alternative hypothesis (H1) is that the average yield is not equal to 12.0 quintals per hectare. First, we calculate the mean (average) yield from the given data: Mean yield = (14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1) / 10 = 12.78 quintals/hectare Next, we calculate the standard deviation of the yields: Standard deviation (SD) = √[(Σ(yi - mean)^2) / (n-1)] where yi is each individual yield and n is the number of observations (10 in this case). SD = √[(1.69 + 0.18 + 1.09 + 3.81 + 1.09 + 0.64 + 1.96 + 0.64 + 0.18 + 1.31) / 9] = √[12.68 / 9] = √1.41 = 1.19 quintals/hectare Now, we calculate the standard error (SE) of the mean: SE = SD / √n = 1.19 / √10 = 0.38 quintals/hectare To conduct the One-Sample T-Test, we calculate the t-value: t = (mean - expected yield) / SE = (12.78 - 12.0) / 0.38 = 2.05 With 9 degrees of freedom (n-1), we look up the critical t-value for a 5% significance level in a t-distribution table. The critical t-value for 9 degrees of freedom at a 5% significance level (two-tailed test) is approximately 2.26. Since our calculated t-value (2.05) is less than the critical t-value (2.26), we fail to reject the null hypothesis. This means that the results conform to the expected yield of 12.0 quintals per hectare at a 5% significance level. In conclusion, the average yield of the new variety green gram, based on the field experiments, does not significantly differ from the expected yield of 12.0 quintals per hectare at a 5% significance level. |
True |
| zhipu | To determine if the results conform to the expectation of a yield of 12.0 quintals per hectare, we can perform a hypothesis test. Specifically, we can use a one-sample t-test to compare the mean yield from the sample to the expected yield. Here are the steps to perform this test: ### Step 1: State the Hypotheses - **Null Hypothesis (H₀):** The mean yield of the new variety is 12.0 quintals per hectare (μ = 12.0). - **Alternative Hypothesis (H₁):** The mean yield of the new variety is not 12.0 quintals per hectare (μ ≠ 12.0). ### Step 2: Choose the Significance Level - The significance level (α) is given as 5% (0.05). ### Step 3: Calculate the Sample Mean and Standard Deviation - Sample yields: 14.3, 12.6, 13.7, 10.9, 13.7, 12.0, 11.4, 12.0, 12.6, 13.1 - Sample size (n) = 10 #### Sample Mean (\(\bar{x}\)): \[ \bar{x} = \frac{\sum x_i}{n} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} = \frac{127.3}{10} = 12.73 \] #### Sample Standard Deviation (s): \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] Calculating each term: \[ (14.3 - 12.73)^2 = 1.1089, \quad (12.6 - 12.73)^2 = 0.0529, \quad (13.7 - 12.73)^2 = 0.9409 \] \[ (10.9 - 12.73)^2 = 3.0129, \quad (13.7 - 12.73)^2 = 0.9409, \quad (12.0 - 12.73)^2 = 0.5329 \] \[ (11.4 - 12.73)^2 = 2.1976, \quad (12.0 - 12.73)^2 = 0.5329, \quad (12.6 - 12.73)^2 = 0.0529 \] \[ (13.1 - 12.73)^2 = 0.1369 \] Sum of squares: \[ \sum (x_i - \bar{x})^2 = 1.1089 + 0.0529 + 0.9409 + 3.0129 + 0.9409 + 0.5329 + 2.1976 + 0.5329 + 0.0529 + 0.1369 = 9.6988 \] \[ s = \sqrt{\frac{9.6988}{9}} = \sqrt{1.0776} \approx 1.038 \] ### Step 4: Calculate the t-Statistic \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{12.73 - 12.0}{1.038 / \sqrt{10}} = \frac{0.73}{0.325} \approx 2.246 \] ### Step 5: Determine the Critical t-Value - Degrees of freedom (df) = n - 1 = 10 - 1 = 9 - For a two-tailed test at α = 0.05 and df = 9, the critical t-value from the t-distribution table is approximately ±2.262. ### Step 6: Compare the t-Statistic to the Critical t-Value - If \(|t| \leq t_{\text{critical}}\), we fail to reject the null hypothesis. - If \(|t| > t_{\text{critical}}\), we reject the null hypothesis. In this case: \[ |t| = 2.246 \quad \text{and} \quad t_{\text{critical}} = 2.262 \] Since \(2.246 < 2.262\), we fail to reject the null hypothesis. ### Conclusion At the 5% significance level, there is not enough evidence to conclude that the mean yield of the new variety of green gram is different from 12.0 quintals per hectare. Therefore, the results conform to the expectation. |
True |
| gpto1 | **Solution:** To determine whether the observed yields conform to the expected yield of 12.0 quintals per hectare at a 5% significance level, we will perform a hypothesis test using the sample data from the 10 farmers' fields. **Step 1: State the Hypotheses** - **Null Hypothesis (H₀):** The true mean yield of the new green gram variety is 12.0 quintals per hectare. \( H_0: \mu = 12.0 \) - **Alternative Hypothesis (H₁):** The true mean yield is not equal to 12.0 quintals per hectare. \( H_1: \mu \ne 12.0 \) This is a two-tailed test since we are checking for any significant difference (either higher or lower) from the expected mean. **Step 2: Collect and Summarize the Sample Data** The yields (in quintals per hectare) from the 10 farmers' fields are: 1. 14.3 2. 12.6 3. 13.7 4. 10.9 5. 13.7 6. 12.0 7. 11.4 8. 12.0 9. 12.6 10. 13.1 **Calculate the Sample Mean (\( \bar{x} \))** \( \bar{x} = \frac{\text{Sum of all sample yields}}{n} \) \( \bar{x} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} \) \( \bar{x} = \frac{126.3}{10} \) \( \bar{x} = 12.63 \) quintals per hectare **Calculate the Sample Standard Deviation (s)** First, calculate each deviation from the mean and square it. | \( x_i \) | \( (x_i - \bar{x}) \) | \( (x_i - \bar{x})^2 \) | |-----------|------------------------|-------------------------| | 14.3 | 14.3 - 12.63 = 1.67 | \( (1.67)^2 = 2.7889 \) | | 12.6 | 12.6 - 12.63 = -0.03 | \( (-0.03)^2 = 0.0009 \) | | 13.7 | 13.7 - 12.63 = 1.07 | \( (1.07)^2 = 1.1449 \) | | 10.9 | 10.9 - 12.63 = -1.73 | \( (-1.73)^2 = 2.9929 \) | | 13.7 | 13.7 - 12.63 = 1.07 | \( (1.07)^2 = 1.1449 \) | | 12.0 | 12.0 - 12.63 = -0.63 | \( (-0.63)^2 = 0.3969 \) | | 11.4 | 11.4 - 12.63 = -1.23 | \( (-1.23)^2 = 1.5129 \) | | 12.0 | 12.0 - 12.63 = -0.63 | \( (-0.63)^2 = 0.3969 \) | | 12.6 | 12.6 - 12.63 = -0.03 | \( (-0.03)^2 = 0.0009 \) | | 13.1 | 13.1 - 12.63 = 0.47 | \( (0.47)^2 = 0.2209 \) | **Sum of Squared Deviations (SSD):** \( \text{SSD} = 2.7889 + 0.0009 + 1.1449 + 2.9929 + 1.1449 + 0.3969 + 1.5129 + 0.3969 + 0.0009 + 0.2209 \) \( \text{SSD} = 10.6012 \) (rounded to four decimal places) **Calculate the Sample Variance (s²):** \( s^2 = \frac{\text{SSD}}{n - 1} = \frac{10.6012}{10 - 1} = \frac{10.6012}{9} = 1.1780 \) **Calculate the Sample Standard Deviation (s):** \( s = \sqrt{s^2} = \sqrt{1.1780} = 1.0854 \) quintals per hectare **Step 3: Calculate the Test Statistic** Since the population standard deviation is unknown and the sample size is small (\( n < 30 \)), we will use the t-test. **Calculate the Standard Error (SE):** \( \text{SE} = \frac{s}{\sqrt{n}} = \frac{1.0854}{\sqrt{10}} = \frac{1.0854}{3.1623} = 0.3435 \) **Calculate the t-statistic:** \( t = \frac{\bar{x} - \mu_0}{\text{SE}} = \frac{12.63 - 12.0}{0.3435} = \frac{0.63}{0.3435} = 1.834 \) **Step 4: Determine the Critical t-value** Degrees of freedom (df): \( \text{df} = n - 1 = 10 - 1 = 9 \) For a two-tailed test at a 5% significance level (\( \alpha = 0.05 \)), we look up the critical t-value from the t-distribution table for \( df = 9 \). \( t_{\text{critical}} = t_{(\alpha/2, \, df)} = t_{(0.025, \,9)} \) From the t-distribution table: \( t_{\text{critical}} = \pm 2.262 \) **Step 5: Make a Decision** - **If \( |t_{\text{calculated}}| > t_{\text{critical}} \)**, we reject the null hypothesis. - **If \( |t_{\text{calculated}}| \leq t_{\text{critical}} \)**, we fail to reject the null hypothesis. Calculated t-statistic: \( |t_{\text{calculated}}| = |1.834| = 1.834 \) Since: \( 1.834 < 2.262 \) We **fail to reject the null hypothesis**. **Step 6: Conclusion** At the 5% significance level, there is no sufficient evidence to conclude that the true mean yield of the new green gram variety is different from 12.0 quintals per hectare. Therefore, the observed yields **conform to the expectation**. **Answer:** Yes, the results conform to the expectation at the 5% significance level. The statistical analysis shows that there is no significant difference between the observed mean yield and the expected yield of 12.0 quintals per hectare. |
True |
| deepseekv3 | To determine whether the results conform to the expected yield of 12.0 quintals per hectare, we can perform a **one-sample t-test**. Here's the step-by-step solution: --- ### **Step 1: Define the Hypotheses** - **Null Hypothesis (\(H_0\))**: The mean yield is equal to 12.0 quintals/hectare (\(µ = 12.0\)). - **Alternative Hypothesis (\(H_1\))**: The mean yield is not equal to 12.0 quintals/hectare (\(µ \neq 12.0\)). This is a two-tailed test. --- ### **Step 2: Calculate the Sample Mean and Standard Deviation** Given the sample data: \[ 14.3, 12.6, 13.7, 10.9, 13.7, 12.0, 11.4, 12.0, 12.6, 13.1 \] 1. **Sample Mean (\(\bar{x}\))**: \[ \bar{x} = \frac{\sum x_i}{n} = \frac{14.3 + 12.6 + 13.7 + 10.9 + 13.7 + 12.0 + 11.4 + 12.0 + 12.6 + 13.1}{10} = \frac{126.3}{10} = 12.63 \text{ quintals/hectare} \] 2. **Sample Standard Deviation (\(s\))**: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] Calculate each \((x_i - \bar{x})^2\): \[ (14.3-12.63)^2 = 2.79, \quad (12.6-12.63)^2 = 0.00, \quad (13.7-12.63)^2 = 1.14, \quad (10.9-12.63)^2 = 2.99, \quad (13.7-12.63)^2 = 1.14, \quad (12.0-12.63)^2 = 0.40, \quad (11.4-12.63)^2 = 1.51, \quad (12.0-12.63)^2 = 0.40, \quad (12.6-12.63)^2 = 0.00, \quad (13.1-12.63)^2 = 0.22 \] Sum of squared deviations: \[ 2.79 + 0.00 + 1.14 + 2.99 + 1.14 + 0.40 + 1.51 + 0.40 + 0.00 + 0.22 = 10.59 \] Sample standard deviation: \[ s = \sqrt{\frac{10.59}{9}} = \sqrt{1.18} = 1.09 \text{ quintals/hectare} \] --- ### **Step 3: Calculate the Test Statistic (t-score)** The formula for the t-score is: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where: - \(\bar{x} = 12.63\) (sample mean), - \(\mu_0 = 12.0\) (expected mean), - \(s = 1.09\) (sample standard deviation), - \(n = 10\) (sample size). Substitute the values: \[ t = \frac{12.63 - 12.0}{1.09 / \sqrt{10}} = \frac{0.63}{0.345} = 1.83 \] --- ### **Step 4: Determine the Critical Value** For a two-tailed t-test with a significance level of 5% (\(\alpha = 0.05\)) and degrees of freedom (\(df = n-1 = 9\)), the critical t-value is: \[ t_{\text{critical}} = \pm 2.262 \] --- ### **Step 5: Compare the Test Statistic with the Critical Value** The calculated t-score (\(t = 1.83\)) lies within the range \([-2.262, 2.262]\). Therefore, we fail to reject the null hypothesis. --- ### **Step 6: Conclusion** At the 5% level of significance, there is **not enough evidence** to reject the null hypothesis. This means the results **conform to the expectation** that the mean yield is 12.0 quintals per hectare. --- ### **Final Answer** The results conform to the expectation of a yield of 12.0 quintals per hectare at the 5% level of significance. |
True |