Math Problem Statement

Students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Assume that a simple random sample has been selected. Use a 0.01 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? 69, 80, 38, 64, 41, 21, 59, 62, 65, 49, 64, 71, 94, 89, 65 Identify the test statistic. Find p-value

Solution

To determine if the students' times are reasonably close to 60 seconds on average, we'll perform a one-sample t-test. This test will help us assess if the sample mean significantly deviates from 60 seconds at a 0.01 significance level.

Here’s the step-by-step solution:

Step 1: Set up Hypotheses

  1. Null Hypothesis (H₀): The mean time is equal to 60 seconds. H0:μ=60H_0: \mu = 60
  2. Alternative Hypothesis (H₁): The mean time is not equal to 60 seconds. H1:μ60H_1: \mu \neq 60

Step 2: Sample Mean and Standard Deviation

Calculate the sample mean xˉ\bar{x} and sample standard deviation ss based on the given times.

  • Sample Data: 69,80,38,64,41,21,59,62,65,49,64,71,94,89,6569, 80, 38, 64, 41, 21, 59, 62, 65, 49, 64, 71, 94, 89, 65
  1. Sample Size (n): n=15n = 15
  2. Sample Mean (xˉ\bar{x}): Compute xˉ\bar{x}.
  3. Sample Standard Deviation (s): Compute ss.

Step 3: Test Statistic

The test statistic tt for a one-sample t-test is calculated by: t=xˉμs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}} where:

  • μ=60\mu = 60 (population mean under the null hypothesis),
  • xˉ\bar{x} is the sample mean,
  • ss is the sample standard deviation,
  • nn is the sample size.

Step 4: Find the p-value

Using the calculated tt-value and degrees of freedom (df=n1=14df = n - 1 = 14), find the p-value for a two-tailed test.

Step 5: Compare p-value to Significance Level

If the p-value is less than 0.01, we reject the null hypothesis, indicating that the sample mean is significantly different from 60 seconds.

I'll calculate these values for you.### Results

  1. Sample Mean (xˉ\bar{x}): 62.0762.07 seconds
  2. Sample Standard Deviation (s): 19.1019.10 seconds
  3. Test Statistic (t): 0.4190.419
  4. p-value: 0.6820.682

Conclusion

Since the p-value 0.6820.682 is much greater than the significance level of 0.010.01, we fail to reject the null hypothesis. This suggests that there is no significant evidence to conclude that the average estimated time by students is different from 60 seconds.

Interpretation

It appears that students are reasonably accurate in estimating one minute, as their mean estimation time does not significantly deviate from 60 seconds.


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Related Questions

  1. What would the outcome be if the significance level were 0.05?
  2. How would the result change if the sample size were larger?
  3. Why is a t-test appropriate here instead of a z-test?
  4. What assumptions are made in this one-sample t-test?
  5. How sensitive is the t-test to outliers in the sample data?

Tip

When dealing with small sample sizes, the t-distribution provides a more reliable estimate than the normal distribution, especially if the population standard deviation is unknown.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
One-Sample T-Test

Formulas

Sample Mean: x̄ = (Σx) / n
Sample Standard Deviation: s = sqrt[(Σ(x - x̄)^2) / (n - 1)]
Test Statistic (t): t = (x̄ - μ) / (s / sqrt(n))

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate