问题


There is a single choice question about high school statistics. Answer the question by replying A, B, C or D.
Question: A farmer who raises hens for egg production wants his eggs to have a mean mass of 56 grams. He is considering the purchase of a different type of hen, so he took a random sample of 18 eggs laid by this type of hen. The distribution of the masses is symmetric and mound-shaped with a mean of 54.1 grams and no outliers. The farmer conducted a t-test to see if there is evidence that the eggs from these hens have a mean mass that is different from 56 g and got a test statistic of t= -1.973. If he uses a 5% significance level, which is the correct conclusion and reason?
A. Because tis more extreme than ±1.96, he should reject the null hypothesis. He has convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
B. Because tis less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
C. Because tis less extreme than the critical value of t for 18 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
D. Because tis more extreme than the critical value of t for 18 degrees of freedom, he should reject the null hypothesis. He has convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
Answer:

答案

模型 模型答案 正确答案
金声玉亮2.0

B. Because tis less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
B
deepseekr1

B. Because t is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.

**Answer:** B
B
gpto1 To answer this question, we need to evaluate whether the calculated test statistic \( t = -1.973 \) is more or less extreme than the critical t-value at the 5% significance level for 17 degrees of freedom (since \( n - 1 = 18 - 1 = 17 \)).

For a two-tailed test at the 5% significance level and 17 degrees of freedom, the critical t-values are approximately \( \pm2.110 \). This means that the rejection region for the null hypothesis consists of t-values less than \(-2.110\) or greater than \(2.110\).

Since the calculated test statistic \( t = -1.973 \) is between \(-2.110\) and \(2.110\), it is **less extreme** than the critical value. Therefore, we **fail to reject** the null hypothesis at the 5% significance level. This means there isn't sufficient evidence to conclude that the mean mass of the eggs is different from 56 grams.

Option B correctly states this conclusion and uses the appropriate degrees of freedom and critical t-values for the test.

**Answer:** B
B
GPT-4o B. Because t is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams. B
llama3.1-70B-Instruct

B
B
llama3.1-405B-Instruct

B. Because tis less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
B
DeepSeek-Coder-V2 A B
Qwen2-72B-Instruct B. Because t is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams. B
baichuan4 C B
doubao C

For a two-tailed test at a 5% significance level with 17 degrees of freedom (n - 1 = 18 - 1 = 17), the critical values are approximately ±2.110. The test statistic t = -1.973 is less extreme than the critical value, so we should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams. And the degrees of freedom is 17 not 18.
B
minimax B. Because t is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams. B
kimi B. Because t is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.

Explanation: The farmer is conducting a one-sample t-test to determine if the mean mass of the eggs from the new type of hen is significantly different from 56 grams. The degrees of freedom for this test are calculated as the sample size minus one, which is 18 - 1 = 17.

The critical value for a two-tailed t-test at a 5% significance level with 17 degrees of freedom can be found using a t-distribution table or a calculator. For a two-tailed test at the 5% level, the critical t-value is approximately ±2.110 (this value can vary slightly depending on the source of the t-distribution table).

Since the calculated t-statistic of -1.973 is less extreme (closer to zero) than the critical value of ±2.110, the farmer does not have enough evidence to reject the null hypothesis that the mean mass of the eggs is equal to 56 grams.
B
wenxinyiyan A

Explanation:
The farmer conducted a t-test with a sample size of 18 (which means 17 degrees of freedom because df = n - 1) to test if the mean mass of eggs from these hens is different from 56 grams. The test statistic he obtained is t = -1.973.

At a 5% significance level, the critical values for a two-tailed test with 17 degrees of freedom are approximately ±1.96 (these are the values that cut off 2.5% of the area in each tail of the t-distribution). Since the absolute value of the test statistic (|t| = 1.973) is greater than the critical value of 1.96, the farmer should reject the null hypothesis.

This means that he has convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams. Therefore, the correct answer is A.
B
zhipu To determine the correct conclusion, we need to compare the test statistic to the critical value of the t-distribution at the given significance level.

For a t-test at a 5% significance level with 17 degrees of freedom (since \( n = 18 \), degrees of freedom \( df = n - 1 = 17 \)), we need to look up the critical value of \( t \) for a two-tailed test.

The critical value of \( t \) for a two-tailed test with 17 degrees of freedom at a 5% significance level is approximately ±2.1098.

Given the test statistic \( t = -1.973 \):
- This test statistic is less extreme than ±2.1098 (i.e., \( -1.973 > -2.1098 \) and \( 1.973 < 2.1098 \)).

Therefore, the correct conclusion is that the test statistic is not extreme enough to reject the null hypothesis.

Option B correctly states this reasoning:
- Because \( t \) is less extreme than the critical value of \( t \) for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.

So, the correct answer is:

B. Because \( t \) is less extreme than the critical value of \( t \) for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
B
gpto1 To answer this question, we need to evaluate whether the calculated test statistic \( t = -1.973 \) is more or less extreme than the critical t-value at the 5% significance level for 17 degrees of freedom (since \( n - 1 = 18 - 1 = 17 \)).

For a two-tailed test at the 5% significance level and 17 degrees of freedom, the critical t-values are approximately \( \pm2.110 \). This means that the rejection region for the null hypothesis consists of t-values less than \(-2.110\) or greater than \(2.110\).

Since the calculated test statistic \( t = -1.973 \) is between \(-2.110\) and \(2.110\), it is **less extreme** than the critical value. Therefore, we **fail to reject** the null hypothesis at the 5% significance level. This means there isn't sufficient evidence to conclude that the mean mass of the eggs is different from 56 grams.

Option B correctly states this conclusion and uses the appropriate degrees of freedom and critical t-values for the test.

**Answer:** B
B
deepseekv3 The correct answer is **B**.

### Explanation:
1. **Test Statistic and Critical Value**:
- The test statistic is \( t = -1.973 \).
- The degrees of freedom (df) for this t-test are \( n - 1 = 18 - 1 = 17 \).
- At a 5% significance level and 17 degrees of freedom, the critical t-value (two-tailed) is approximately \( \pm 2.110 \).

2. **Comparison**:
- The test statistic \( t = -1.973 \) is **less extreme** than the critical value \( \pm 2.110 \).

3. **Conclusion**:
- Since the test statistic does not fall in the rejection region (i.e., it is not more extreme than the critical value), the farmer **should not reject the null hypothesis**.
- There is not enough evidence at the 5% significance level to conclude that the mean mass of eggs from these hens is different from 56 grams.

Thus, the correct answer is:

**B. Because t is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.**
B