[Solved] Identify the error in this code snippet for calculating a 90% confidence interval:from scipy.stats import t sample_mean = 10 sample_std = 2 n = 25 se = sample_std / n interval = t.interval(0.... | SciPy

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Identify the error in this code snippet for calculating a 90% confidence interval:

from scipy.stats import t
sample_mean = 10
sample_std = 2
n = 25
se = sample_std / n
interval = t.interval(0.90, n-1, loc=sample_mean, scale=se)
print(interval)

ADegrees of freedom should be n, not n-1

BWrong confidence level value

CStandard error calculation is incorrect

Dt.interval function does not exist

Step-by-Step Solution

Solution:

Step 1: Check standard error calculation
Standard error should be sample_std divided by sqrt(n), not by n.
Step 2: Verify other parts
Confidence level 0.90 and degrees of freedom n-1 are correct; t.interval exists.
Final Answer:
Standard error calculation is incorrect -> Option C
Quick Check:
SE = std / sqrt(n), not std / n [OK]

Quick Trick: SE = std / sqrt(n), not std / n [OK]

Common Mistakes:

Dividing std by n instead of sqrt(n)
Confusing degrees of freedom
Using wrong confidence level format

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Identify the error in this code snippet for calculating a 90% confidence interval:

Step 1: Check standard error calculation

Step 2: Verify other parts

Final Answer:

Quick Check:

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