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You have a dataset with 100 measurements and want a 99% confidence interval for the mean. Which code correctly computes it using scipy?

hard📝 Application Q15 of 15
SciPy - Curve Fitting and Regression
You have a dataset with 100 measurements and want a 99% confidence interval for the mean. Which code correctly computes it using scipy?
Afrom scipy.stats import t import numpy as np data = np.random.randn(100) mean = np.mean(data) se = np.std(data) / 100 interval = t.interval(0.99, 100, loc=mean, scale=se) print(interval)
Bfrom scipy.stats import t import numpy as np data = np.random.randn(100) mean = np.mean(data) se = np.std(data, ddof=1) / np.sqrt(100) interval = t.interval(0.99, 99, loc=mean, scale=se) print(interval)
Cfrom scipy.stats import t import numpy as np data = np.random.randn(100) mean = np.mean(data) se = np.std(data, ddof=1) / np.sqrt(100) interval = t.interval(0.95, 99, loc=mean, scale=se) print(interval)
Dfrom scipy.stats import t import numpy as np data = np.random.randn(100) mean = np.mean(data) se = np.std(data, ddof=1) / 100 interval = t.interval(0.99, 99, loc=mean, scale=se) print(interval)
Step-by-Step Solution
Solution:
  1. Step 1: Check standard error calculation

    Standard error must be sample std dev with ddof=1 divided by sqrt(n), which is 100 here.
  2. Step 2: Check confidence level and degrees of freedom

    99% confidence means 0.99; degrees of freedom = n-1 = 99.
  3. Step 3: Verify code correctness

    from scipy.stats import t import numpy as np data = np.random.randn(100) mean = np.mean(data) se = np.std(data, ddof=1) / np.sqrt(100) interval = t.interval(0.99, 99, loc=mean, scale=se) print(interval) correctly uses ddof=1, sqrt(100), 0.99 confidence, and 99 degrees of freedom.
  4. Final Answer:

    The code with ddof=1, /np.sqrt(100), 0.99 confidence, df=99 -> Option B
  5. Quick Check:

    Use ddof=1, sqrt(n), 0.99 confidence, df=n-1 [OK]
Quick Trick: Use ddof=1 and sqrt(n) for SE; df = n-1 [OK]
Common Mistakes:
  • Using population std dev (ddof=0)
  • Dividing std by n instead of sqrt(n)
  • Wrong confidence level or degrees of freedom

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