Confidence intervals on parameters in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Calculate 95% confidence interval for mean with scipy

Given the sample data below, what is the 95% confidence interval for the mean using scipy.stats.t.interval?

SciPy

import numpy as np
from scipy import stats

data = np.array([5, 7, 8, 6, 9, 10, 7, 6])

mean = np.mean(data)
sem = stats.sem(data)
conf_int = stats.t.interval(0.95, len(data)-1, loc=mean, scale=sem)
print(conf_int)

A (5.35, 8.90)

B (5.50, 8.75)

C (5.60, 8.60)

D (5.10, 9.20)

💡 Hint

Use stats.sem to get the standard error of the mean, then use stats.t.interval with degrees of freedom = sample size - 1.

The 95% confidence interval for the mean is calculated using the t-distribution because the sample size is small. The standard error of the mean (SEM) is computed, then stats.t.interval returns the interval around the sample mean.

❓ data_output

intermediate

Interpret confidence interval output for regression coefficient

After fitting a linear regression, you get the 95% confidence interval for the slope as (1.2, 3.4). What does this interval mean?

A The true slope is between 1.2 and 3.4 with 95% confidence.

B 95% of the data points fall between 1.2 and 3.4.

C The slope will be exactly 2.3 in 95% of future samples.

D The slope is either 1.2 or 3.4 with 95% probability.

💡 Hint

Confidence intervals estimate where the true parameter lies with a certain confidence level.

The confidence interval means we are 95% confident that the true slope parameter lies between 1.2 and 3.4. It does not describe data points or probabilities of exact values.

🔧 Debug

advanced

Identify error in confidence interval calculation code

What error will this code raise?

import numpy as np
from scipy import stats

data = np.array([2, 4, 6, 8, 10])
mean = np.mean(data)
sem = stats.sem(data)
conf_int = stats.t.interval(0.95, len(data), loc=mean, scale=sem)
print(conf_int)

SciPy

import numpy as np
from scipy import stats

data = np.array([2, 4, 6, 8, 10])
mean = np.mean(data)
sem = stats.sem(data)
conf_int = stats.t.interval(0.95, len(data), loc=mean, scale=sem)
print(conf_int)

A IndexError: index out of range

B ValueError: degrees of freedom <= 0 for slice

C No error, prints confidence interval

D TypeError: unsupported operand type(s)

💡 Hint

Check the degrees of freedom parameter for stats.t.interval.

The degrees of freedom should be sample size minus 1. Using len(data) causes degrees of freedom to be 5, which is valid, so no error is raised. The error would occur if degrees of freedom were 0 or negative.

🚀 Application

advanced

Calculate confidence interval for variance using chi-square distribution

You have a sample variance of 4.0 from 10 observations. Using the chi-square distribution, what is the 95% confidence interval for the population variance?

SciPy

import scipy.stats as stats

n = 10
sample_var = 4.0
alpha = 0.05

lower = (n - 1) * sample_var / stats.chi2.ppf(1 - alpha/2, n - 1)
upper = (n - 1) * sample_var / stats.chi2.ppf(alpha/2, n - 1)
print((lower, upper))

A (2.10, 10.50)

B (1.20, 13.00)

C (1.80, 11.20)

D (1.56, 12.03)

💡 Hint

Use chi-square percent point function (ppf) with degrees of freedom = n - 1.

The confidence interval for variance uses chi-square distribution quantiles. The formula is ((n-1)*s² / chi2_upper, (n-1)*s² / chi2_lower).

🧠 Conceptual

expert

Why use t-distribution for confidence intervals on small samples?

Why is the t-distribution preferred over the normal distribution when calculating confidence intervals for the mean with small sample sizes?

A Because the normal distribution cannot be used for any sample size below 30.

B Because the t-distribution has heavier tails to model outliers better.

C Because the sample variance is unknown and estimated, increasing uncertainty.

D Because the t-distribution is symmetric while the normal distribution is not.

💡 Hint

Think about what changes when the sample size is small and variance is estimated.

When the sample size is small, the population variance is unknown and must be estimated from the sample. This adds uncertainty, so the t-distribution, which accounts for this extra variability, is used instead of the normal distribution.