Confidence intervals on parameters in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the mean of the data array.

SciPy

import numpy as np

data = np.array([5, 7, 8, 9, 10])
mean_value = np.[1](data)
print(mean_value)

Drag options to blanks, or click blank then click option'

Amedian

Bmean

Csum

Dstd

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.median instead of np.mean

Using np.sum which adds all values instead of averaging

The np.mean() function calculates the average (mean) of the array elements.

2fill in blank

medium

Complete the code to calculate the standard error of the mean (SEM) for the data.

SciPy

import numpy as np

data = np.array([5, 7, 8, 9, 10])
sem = np.std(data, ddof=1) / np.sqrt([1])
print(sem)

Drag options to blanks, or click blank then click option'

Alen(data) - 1

Bnp.sum(data)

Cnp.mean(data)

Dlen(data)

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(data) - 1 instead of len(data)

Using np.mean(data) or np.sum(data) which are unrelated

The SEM is the standard deviation divided by the square root of the sample size, which is len(data).

3fill in blank

hard

Fix the error in the code to calculate a 95% confidence interval for the mean using scipy.stats.

SciPy

from scipy import stats
import numpy as np

data = np.array([5, 7, 8, 9, 10])
confidence = 0.95
mean = np.mean(data)
sem = stats.sem(data, ddof=1)
margin = sem * stats.t.ppf((1 + confidence) / 2, df=[1])
ci_lower = mean - margin
ci_upper = mean + margin
print(ci_lower, ci_upper)

Drag options to blanks, or click blank then click option'

Alen(data)

Blen(data) + 1

Clen(data) - 1

Dlen(data) - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(data) instead of len(data) - 1 for degrees of freedom

Using len(data) + 1 or len(data) - 2 which are incorrect

The degrees of freedom (df) for the t-distribution is the sample size minus 1, so len(data) - 1.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths only if the length is greater than 3.

SciPy

words = ['apple', 'bat', 'carrot', 'dog', 'elephant']
lengths = {word: [1] for word in words if [2]
print(lengths)

Drag options to blanks, or click blank then click option'

Alen(word)

Bword

Clen(word) > 3

Dword > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length

Checking if the word string is greater than 3 instead of length

The dictionary comprehension maps each word to its length using len(word). The condition filters words with length greater than 3 using len(word) > 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their lengths only if the length is greater than 4.

SciPy

words = ['apple', 'bat', 'carrot', 'dog', 'elephant']
lengths = { [1]: [2] for word in words if [3] }
print(lengths)

Drag options to blanks, or click blank then click option'

Aword.upper()

Blen(word)

Clen(word) > 4

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using word.lower() instead of word.upper() for keys

Using word instead of len(word) for values

Filtering with length greater than 3 instead of 4

The dictionary comprehension uses word.upper() as keys, maps to len(word), and filters words with length greater than 4 using len(word) > 4.

Practice

1. What does a confidence interval represent in statistics?

easy

A. A range of values likely containing the true parameter

B. The exact value of the parameter

C. The average of the sample data

D. The maximum value observed in the data

Solution

1. Step 1: Understand the meaning of confidence interval

   A confidence interval gives a range where the true parameter is likely to be found, not a single exact value.

2. Step 2: Compare options with definition

   Only A range of values likely containing the true parameter correctly describes this range; others describe different concepts.

3. Final Answer:

   A range of values likely containing the true parameter -> Option A

4. Quick Check:

   Confidence interval = range of likely parameter values [OK]

Hint: Confidence interval = range, not exact value [OK]

Common Mistakes:

• Thinking it gives exact parameter value
• Confusing with sample mean
• Assuming it shows data maximum

2. Which of the following is the correct way to import the function to calculate confidence intervals from scipy?

easy

A. from scipy.stats import t

B. import scipy.confidence as conf

C. from scipy import confidence_interval

D. import scipy.stats.confidence

Solution

1. Step 1: Recall scipy.stats module usage

   The t-distribution and its interval function are in scipy.stats, imported as 'from scipy.stats import t'.

2. Step 2: Check other options

   Other imports do not exist or are incorrect syntax.

3. Final Answer:

   from scipy.stats import t -> Option A

4. Quick Check:

   Correct import for t interval = from scipy.stats import t [OK]

Hint: Use 'from scipy.stats import t' for confidence intervals [OK]

Common Mistakes:

• Trying to import non-existent modules
• Using wrong import syntax
• Confusing function location

3. What is the output of the following code?

import numpy as np
from scipy.stats import t

data = np.array([5, 7, 8, 6, 9])
mean = np.mean(data)
se = np.std(data, ddof=1) / np.sqrt(len(data))
interval = t.interval(0.95, len(data)-1, loc=mean, scale=se)
print(tuple(round(x, 2) for x in interval))

medium

A. (5.00, 9.00)

B. (4.50, 9.30)

C. (6.00, 7.00)

D. (5.04, 8.96)

Solution

1. Step 1: Calculate mean and standard error

   Mean = (5+7+8+6+9)/5 = 7.0; sample std dev ≈ 1.58; SE = 1.58 / sqrt(5) ≈ 0.71.

2. Step 2: Calculate 95% confidence interval using t-distribution

   Degrees of freedom = 4; t critical ≈ 2.776; interval = mean ± t * SE = 7.0 ± 2.776*0.71 ≈ (5.04, 8.96).

3. Final Answer:

   (5.04, 8.96) -> Option D

4. Quick Check:

   Mean ± t*SE = (5.04, 8.96) [OK]

Hint: Calculate mean, SE, then apply t.interval [OK]

Common Mistakes:

• Using population std dev instead of sample
• Wrong degrees of freedom
• Rounding errors

4. 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)

medium

A. Degrees of freedom should be n, not n-1

B. Wrong confidence level value

C. Standard error calculation is incorrect

D. t.interval function does not exist

Solution

1. Step 1: Check standard error calculation

   Standard error should be sample_std divided by sqrt(n), not by n.

2. Step 2: Verify other parts

   Confidence level 0.90 and degrees of freedom n-1 are correct; t.interval exists.

3. Final Answer:

   Standard error calculation is incorrect -> Option C

4. Quick Check:

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

Hint: 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

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

hard

A. from 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)

B. 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)

C. 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(0.95, 99, loc=mean, scale=se)
print(interval)

D. from 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)

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]

Hint: 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