Question 1168567: A teacher expected students to take about 20 mins on their online exam. She noticed that a lot of her students were taking less time and she is wondering if there is a systematic difference or if what she is seeing is just due to chance. Did her class of 48 students differ significantly from her typical average of 20 mins?
Question:
1. Statistical Hypotheses?
2. reject the null hypothesis at the .05 level of significance if t ≥ __
or t ≤ __ given __ df.
3. what is your decision regarding whether she should reject or accept the null hypothesis.
4. How do you interpret your decision?
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! Let's break down this hypothesis test step-by-step.
**1. Statistical Hypotheses**
* **Null Hypothesis (H₀):** The mean time taken by the students is 20 minutes (μ = 20).
* **Alternative Hypothesis (H₁):** The mean time taken by the students is different from 20 minutes (μ ≠ 20).
This is a two-tailed test because we are testing if the time is *different* from 20 minutes, not specifically less than or greater than.
**2. Reject the Null Hypothesis**
* Significance level (α) = 0.05
* Sample size (n) = 48
* Degrees of freedom (df) = n - 1 = 48 - 1 = 47
Since the sample size is large (n > 30), we can use the t-distribution or approximate it with the z-distribution. However, it's safer to use the t-distribution with df = 47.
* Using a t-table or calculator:
* t(0.025, 47) ≈ ±2.012 (This is the critical t-value for a two-tailed test with α = 0.05 and df = 47).
* Reject H₀ if t ≥ 2.012 or t ≤ -2.012.
**3. Decision (We need More Information to Make a Decision)**
To make a decision, we need the following information:
* **Sample mean (x̄):** The average time her class actually took.
* **Sample standard deviation (s):** The standard deviation of the times her class took.
Without this information, we cannot calculate the t-statistic and make a decision.
**Example Scenario (to illustrate the decision-making process):**
Let's assume:
* Sample mean (x̄) = 18 minutes
* Sample standard deviation (s) = 5 minutes
Now, we can calculate the t-statistic:
* t = (x̄ - μ) / (s / √n)
* t = (18 - 20) / (5 / √48)
* t = -2 / (5 / 6.928)
* t = -2 / 0.7216
* t ≈ -2.772
In this example, t = -2.772. Since -2.772 ≤ -2.012, we would reject the null hypothesis.
**4. Interpretation (Based on the Example Scenario)**
If we used the example data, the interpretation would be:
* The results of the t-test indicate that there is a statistically significant difference in the average time taken by the students compared to the teacher's expected 20 minutes.
* Specifically, the students took significantly less time than expected (18 minutes on average in our example).
* This suggests that the teacher's observation of students taking less time is not likely due to chance and that there is a systematic difference.
**Important Note:**
* The actual decision and interpretation depend on the sample mean and standard deviation, which were not provided in the original question.
* If you provide the sample mean and standard deviation, I can give you a more specific answer.
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