SOLUTION: A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.10 significance level to test
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Question 1168843: A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?
70 84 42 65 41 24 58 63 69 49 61 72 90 91 67
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
Need t find
test statstic and p-value
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this hypothesis test step-by-step.
**1. Define the Hypotheses**
* **Null Hypothesis (H₀):** The mean estimated time is equal to 60 seconds.
* H₀: µ = 60
* **Alternative Hypothesis (H₁):** The mean estimated time is not equal to 60 seconds.
* H₁: µ ≠ 60 (This is a two-tailed test)
**2. Set the Significance Level**
* α = 0.10
**3. Calculate the Sample Statistics**
* Data: 70, 84, 42, 65, 41, 24, 58, 63, 69, 49, 61, 72, 90, 91, 67
* Sample size (n) = 15
* **Sample Mean (x̄):**
* Sum of the data: 70 + 84 + 42 + 65 + 41 + 24 + 58 + 63 + 69 + 49 + 61 + 72 + 90 + 91 + 67 = 946
* x̄ = 946 / 15 ≈ 63.0667
* **Sample Standard Deviation (s):**
* Using a calculator or statistical software, we find s ≈ 20.3061
**4. Calculate the Test Statistic (t)**
* Since the population standard deviation is unknown and the sample size is small (n < 30), we use a t-test.
* Formula: t = (x̄ - µ) / (s / √n)
* Calculation:
* t = (63.0667 - 60) / (20.3061 / √15)
* t ≈ 3.0667 / (20.3061 / 3.87298)
* t ≈ 3.0667 / 5.2429
* t ≈ 0.585
**5. Calculate the Degrees of Freedom (df)**
* df = n - 1 = 15 - 1 = 14
**6. Calculate the P-value**
* Using a t-distribution table or statistical software, we find the p-value for a two-tailed t-test with t ≈ 0.585 and df = 14.
* Using a t-table or calculator, the p-value is approximately 0.567.
**7. Make a Decision**
* Compare the p-value (0.567) to the significance level (0.10).
* Since 0.567 > 0.10, we fail to reject the null hypothesis.
**8. Draw a Conclusion**
* There is not enough evidence to reject the claim that the mean estimated time is equal to 60 seconds at the 0.10 significance level.
* It appears that students are reasonably good at estimating one minute.
**Summary**
* **Test Statistic (t):** ≈ 0.585
* **P-value:** ≈ 0.567
* **Decision:** Fail to reject the null hypothesis.
* **Conclusion:** The data does not provide sufficient evidence to conclude that the mean estimated time is different from 60 seconds.
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