T-test

T-test

Q1. Green Gram Yield

Traditionally it is known that a green gram cultivation yields 12.0 quintals per hectare on an average.

To increase crop yields, scientists have developed a new variety of green grams, that can supposedly produce more than the expected average yield of 12 quintals per hectare.

To test the same, this variety of green grams was tested on 10 randomly selected farmer's fields.

The yield (quintals/hectare) was recorded as: [14.3,12.6,13.7,10.9,13.7,12.0,11.4,12.0,12.6,13.1]

With a 5% significance level, can we conclude that the average yield of this variety of green gra

ms is more than the expected yield (12 quintals/hectare)?

Perform an appropriate test and choose the correct option below :

A.    p_value = 0.033, Average sample yield is same as expected

B.    p_value = 0.049, Average sample yield is same as expected

C.    p_value = 0.049, Average sample yield is greater than expected

D.    p_value = 0.033, Average sample yield is greater than expected

 

 

Correct Option: p_value = 0.049, Average sample yield is greater than expected

Explanation:
Based on the question, we define our hypothesis as:

• H0 : This variety of green gram will Yield 12 quintals per hectare (Yield = 12)
• Ha : This variety of green gram will Yield more than 12 quintals per hectare (Yield > 12)

We can solve this using One sample T-test

Since we will be performing Right Tail Test, we will set alternative = "greater"

Code:

from scipy.stats import ttest_1samp

 

yield_data = [14.3,12.6,13.7,10.9,13.7,12.0,11.4,12.0,12.6,13.1]

alpha = 0.05  #5% significance level

 

tstat, pvalue = ttest_1samp(yield_data, 12.0, alternative = "greater")

print(pvalue)

 

if pvalue < alpha:

  print('Reject H0 ; Yield will be more than 12 quintals per hectare')

else:

  print('Fail to reject H0 ; Yield will be 12 quintals per hectare')

Output:
0.04979938002326665
Reject H0 ; Yield will be more than 12 quintals per hectare

 

Q2. Gym body fat percentage

Samples of Body fat percentages of few gym going men and women are recorded.

 

men = [13.3, 6.0, 20.0, 8.0, 14.0, 19.0, 18.0, 25.0, 16.0, 24.0, 15.0, 1.0, 15.0]

women = [22.0, 16.0, 21.7, 21.0, 30.0, 26.0, 12.0, 23.2, 28.0, 23.0]

Perform an appropriate test to check if the mean body fat percentage of men and women is statistically different.

 

Assume the significance level to be 5%.

 

Choose the correct option below :

A.    p_value = 0.03, Body fat percentage of gym going men and women are same

B.    p_value = 0.01, Body fat percentage of gym going men and women are same

C.    p_value = 0.03, Body fat percentage of gym going men and women are statistically different

D.    p_value = 0.01, Body fat percentage of gym going men and women are statistically different

 

Correct Option: p_value = 0.01, Body fat percentage of gym going men and women are statistically different

Explanation:
Based on the given problem, we define our hypothesis as:

• H0 : Body fat percentage of men and women are same μ1​=μ2​
• Ha : Body fat percentage of men and women are different μ1​​=μ2​

Since we need to compare two different samples, whose sample size is small (<30), we will use Two Sample T-test

Further, in order to test for the proposed alternate hypothesis, we will have to perform a Two Tailed Test.

Code:

from scipy.stats import ttest_ind

 

# Given samples of gym-going men and women body fat percentages

men = [13.3, 6.0, 20.0, 8.0, 14.0, 19.0, 18.0, 25.0, 16.0, 24.0, 15.0, 1.0, 15.0]

women = [22.0, 16.0, 21.7, 21.0, 30.0, 26.0, 12.0, 23.2, 28.0, 23.0]

alpha = 0.05 # Significance level

 

# Performing 2 Sample Two-Tailed T-test

t_stats, pvalue = ttest_ind(men, women, alternative ='two-sided')

print(pvalue)

 

if pvalue < alpha:

  print('Reject H0 ; Body fat percentage of men and women are different')

else:

  print('Fail to reject H0 ; Body fat percentage of men and women are the same')

Output:
0.010730607904197957
Reject H0 ; Body fat percentage of men and women are different

Q3. Quality assurance

The quality assurance department claims that on average the non-fat milk contains more than 190 mg of Calcium per 500 ml packet.

To check this claim 45 packets of milk are collected and the content of calcium is recorded.

Perform an appropriate test to check the claim with a 90% confidence level.

data = [193, 321, 222, 158, 176, 149, 154, 223, 233, 177, 280, 244, 138, 210, 167, 129, 254, 167, 194, 191, 128, 191, 144, 184, 330, 216, 212, 142, 216, 197, 231, 133, 205, 192, 195, 243, 224, 137, 234, 171, 176, 249, 222, 234, 191]

Note: Round off the answer to four decimal places.

 

 

A.    Test statistic: 1.3689 , Reject null hypothesis

B.    Test statistic: 1.3689 , Fail to reject null hypothesis

C.    Test statistic: 1.2851, Reject null hypothesis

D.    Test statistic: 1.2851 , Fail to reject null hypothesis

 

Correct option: Test statistic: 1.3689 , Reject null hypothesis

Explanation:
Since we need to test the claim made by company, we define our hypothesis as:

• H0: µ <= 190
• H1: µ > 190 (Right-tailed test)

We perform one sample t-test.


Solution Approach code:

from scipy.stats import ttest_1samp
data = pd.Series([193, 321, 222, 158, 176, 149, 154, 223, 233, 177, 280, 244, 138, 210, 167, 129, 254, 167, 194, 191, 128, 191, 144, 184, 330, 216, 212, 142, 216, 197, 231, 133, 205, 192, 195, 243, 224, 137, 234, 171, 176, 249, 222, 234, 191])
print("Observed sample mean = ", round(data.mean(), 2))
t_stat, p_value = ttest_1samp(data, popmean=190, alternative="greater")
print("Test statistic = ", round(t_stat,4))
print("P-value = ", round(p_value,4))
if p_value < 0.10:
 print("Reject H0")
else:
 print("Fail to reject H0")

Output:

Observed sample mean = 199.49
Test statistic = 1.3689
P-value = 0.089
Reject H₀


Therefore, we conclude that the claim that on average the non-fat milk contains more than 190 mg of Calcium per 500 ml packet.

Another way to find the T-statistic is using the formula: t = n​s​x−μ​

 

Q4. Coaching class

There are 8 females and 12 males in a coaching class.

After a practice test, the coach wants to know whether the average score of females is greater than the average score of males.

Given data describes the scores of females and males in his class.

female_scores=[25,30,45,49,47,35,32,42]



male_scores=[45,47,25,22,29,32,27,28,40,49,50,33]

Use an appropriate test to check whether the assumption of the coach is significant or not, at a 2% significance level?

A.    P_value = 0.580, There is significant evidence that the average score of females is greater than the average score of males.

B.    P_value = 0.285, There is no significant evidence that the average score of females is greater than the average score of males.

C.    P_value = 0.285, There is significant evidence that the average score of females is greater than the average score of males.

D.    P_value = 0.580, There is no significant evidence that the average score of females is greater than the average score of males.

 

Correct option: P_value = 0.285, There is no significant evidence that the average score of females is greater than the average score of males.

Explanation:

Based on the given problem, we define our hypothesis as:

• H₀: μ₁ ≤ μ₂, i.e., the average score of females is not greater than the average score of males.
• H₁: μ₁ > μ₂ i.e. the average score of females is greater than the average score of males.
  Based on these proposed hypothesis, we will need to perform Right-tailed test in order to test for the alternate hypothesis.
  
  Code:

import pandas as pd

from scipy.stats import ttest_ind

 

female_scores=pd.Series([25,30,45,49,47,35,32,42])

male_scores=pd.Series([45,47,25,22,29,32,27,28,40,49,50,33])

 

t_stat, p_value = ttest_ind(female_scores, male_scores, alternative="greater")

print("Test statistic = ", round(t_stat,3))

print("P_value =",p_value )

 

alpha=0.02

if p_value < alpha:

    print("Reject the null hypothesis. There is significant evidence that the average score of females is greater than the average score of males.")

else:

    print("Fail to reject the null hypothesis. There is no significant evidence that the average score of females is greater than the average score of males.")

Output:
Test statistic = 0.58
P_value = 0.2847023809445894
Fail to reject the null hypothesis. There is no significant evidence that the average score of females is greater than the average score of males.

 

 

Q5. Effectiveness on yield

An experiment was performed to compare the effectiveness of Ammonium Chloride and urea on the grain yield (in quintal per hectare) and the results are given in the arrays below:

Ammonium_chloride = [13.4, 10.9, 11.2, 11.8, 14, 15.3, 14.2, 12.6, 17, 16.2, 16.5, 15.7]

Urea = [12, 11.7, 10.7, 11.2, 14.8, 14.4, 13.9, 13.7, 16.9, 16, 15.6, 16]

Conduct an appropriate test to compare the same with a 95% confidence level and choose the appropriate option below.

A.    p_val=0.002, The effect of ammonium chloride and urea on grain yield is not equal

B.    p_val=0.855, The effect of ammonium chloride and urea on grain yield is equal

C.    p_val=0.855,The effect of ammonium chloride and urea on grain yield is not equal

D.    p_val=0.002, The effect of ammonium chloride and urea on grain yield is equal

Correct option: p_val=0.855, its greater than alpha, therefore the null hypothesis that effects are equal is accepted.

Explanation:

We define our hypothesis as:

• H₀ = the effect of ammonium chloride and urea on grain yield is equal.
• H₁ = the effect of ammonium chloride and urea on grain yield is not equal.

Since we need to compare two different samples, whose sample size is small (<30), we will use Two Sample T-test

Further, in order to test for the proposed alternate hypothesis, we will have to perform a Two Tailed Test.

Code:

from scipy.stats import ttest_ind

 

# Given samples of farm yields with following chemicals

ammonium_chloride = [13.4, 10.9, 11.2, 11.8, 14, 15.3, 14.2, 12.6, 17, 16.2, 16.5, 15.7]

urea = [12, 11.7, 10.7, 11.2, 14.8, 14.4, 13.9, 13.7, 16.9, 16, 15.6, 16]

alpha = 0.05 # Significance level

 

# Performing 2 Sample Two-Tailed T-test

t_stats, pvalue = ttest_ind(ammonium_chloride, urea, alternative ='two-sided')

print(pvalue)

 

if pvalue < alpha:

  print('Reject H0 ; The effect of ammonium chloride and urea on grain yield is not equal')

else:

  print('Fail to reject H0 ; The effect of ammonium chloride and urea on grain yield is equal')

Output:
0.8551954147800473
Fail to reject H0 ; The effect of ammonium chloride and urea on grain yield is equal

Q6. Zumba trainer

 

The Zumba trainer claims to the customers, that their new dance routine helps to reduce more weight.

Weight of 8 people were recorded before and after following the new Zumba training for a month:

wt_before = [85, 74, 63.5, 69.4, 71.6, 65,90,78]



wt_after = [82, 71, 64, 65.2, 67.8, 64.7,95,77]

Test the trainer's claim with 90% confidence. Further, what would be the pvalue?

A.    P value: 0.854, Customers did not reduce their weight

B.    P value: 0.145, Customers did not reduce their weight

C.    P value: 0.854, Customers have reduced their weight

D.    P value: 0.145, Customers have reduced their weight

 

Correct option: P value: 0.145, Customers did not reduce their weight

Explanation:
Based on the given problem, we define our hypothesis as:

• H₀: Customers did not reduce their weight after 1 month of new Zumba routine μBefore​=μAfter​
• H₁: Customers have reduced their weight after 1 month of new Zumba routine μBefore​>μAfter​

The nature of data here is in the form of “Before” and “After” the new Zumba dance routine. Hence we will use Paired T-test using ttest_rel()

Further, in order to test our proposed alternate hypothesis, we would have to perform Right Tail test

Code:

from scipy.stats import ttest_rel

 

# Given samples of weight before and after 1 month of following new Zumba routine

wt_before = [85, 74, 63.5, 69.4, 71.6, 65,90,78]

wt_after = [82, 71, 64, 65.2, 67.8, 64.7,95,77]

 

alpha = 0.10 # Significance level

 

# Performing Paired Right-Tailed T-test

t_stat, pvalue = ttest_rel(wt_before, wt_after, alternative="greater" )

print('Test Statistic:', t_stat)

print('P value:', pvalue)

 

if pvalue < alpha:

  print('Reject H0 ; Customers have reduced their weight after 1 month of new Zumba routine')

else:

  print('Fail to reject H0 ; Customers did not reduce their weight after 1 month of new Zumba routine')

Output:
Test Statistic: 1.1421853793555032
P value: 0.14546808501326386
Fail to reject H0 ; Customers did not reduce their weight after 1 month of new Zumba routine

Q7. The correct test

A certain company decided to roll out a new training regime for its employees.

To test which regime (old or new) would be preferred by the employees, they made 5 employees (who had earlier cleared the old regime) take part in the new training regime, and then score them both, out of 100.

Which of the following statistical procedures would be most appropriate to test the claim that employee overall scores are the same in both training regimes?

A.    Two-tailed two-sample paired/dependent t-test of means

B.    Two-tailed two-sample independent t-test of means

C.    Two-tailed two-sample independent z-test of means

D.    One-tailed two-sample z-test of means

E.     One-tailed two-sample paired/dependent t-test of means

 

Correct option: Two-tailed two-sample paired/dependent t-test of means

Explanation:

Null and alternative hypotheses:

• Null hypothesis (H₀): The mean score for the new training regime is equal to the mean score for the old training regime. µ_new = µ_old.
• Alternative hypothesis (H₁): The mean score for the new training regime is different from the mean score for the old training regime. µ_new ≠ µ_old.

We know that two sample test is done when we are comparing the mean of two groups with each other. Here the condition for this satisfies.

Similarly, two sided tailed test is used when we evaluate on both sides, and here again, the condition satisfies to evaluate both the regimes.

And finally, since the same sample group takes part in both the before and after regimes, we can conclude that we need a Two-tailed two-sample dependent/paired t-test of means.

Q8. Test the Training Program

You are appointed as a Data Analyst for a training program deployed by the Government of India.

The participants’ skills were tested before and after the training using some metrics on a scale of 10.

before = [2.45, 0.69, 1.80, 2.80, 0.07, 1.67, 2.93, 0.47, 1.45, 1.34]   



after = [7.71, 2.17, 5.65, 8.79, 0.23, 5.23, 9.19, 1.49, 4.56, 4.20] 

Conduct an appropriate test to assess a statistically significant increase in the average skill score after the training program, and then answer the below questions accordingly.

Note: Perform the test at alpha = 5%.

A.    T statistic = -5.11

B.    There is no effect of the training.

C.    There is a positive effect of the training.

D.    We fail to reject the null hypothesis.

E.     We reject the null hypothesis.

 

Correct options:

• T statistic = -5.11
• There is a positive effect of the training.
• We reject the null hypothesis.

Explanation:

We propose the Hypothesis as follows:

• H₀: No improvement in skills after training, i.e. μ1​=μ2​
• H_a: Positive effect / improvement in skills after training, i.e. μ1​<μ2​

We can test this using Paired T-test

Code:

import scipy.stats as stats

 

# Sample of participants’ skills tested before and after the training

before= [2.45, 0.69, 1.80, 2.80, 0.07, 1.67, 2.93, 0.47, 1.45, 1.34]

after = [7.71, 2.17, 5.65, 8.79, 0.23, 5.23, 9.19, 1.49, 4.56, 4.20]

 

before_mean = np.mean(np.array(before))

after_mean = np.mean(np.array(after))

 

print("before mean:",before_mean)

print("after mean:",after_mean)

 

# Performing the paired t-test

t_obs, p = stats.ttest_rel(before, after, alternative="less")

 

print(" T statistic = ", round(t_obs,2))

print(" p-value = ", round(p,4))

 

if(p < 0.05):

  print("Since, p-value < alpha, we reject the null hypothesis. Positive effect / improvement in skills after training")

else:

  print("Since, p-value > alpha, we fail to reject the null hypothesis. No effect / improvement in skills after training")

Output:
before mean: 1.567
after mean: 4.922000000000001
T statistic = -5.11
p-value = 0.0003
Since, p-value < alpha, we reject the null hypothesis. Positive effect / improvement in skills after training.

 

 

Q1. Average British man height

The average British man is 175.3 cm tall. A survey recorded the heights of 10 UK men and we want to know whether the mean of the sample is different from the population mean.

survey_height = [177.3, 182.7, 169.6, 176.3, 180.3, 179.4, 178.5, 177.2, 181.8, 176.5]

Perform an appropriate test and choose the correct option below, that we can conclude with a 5% significance.

A.    p_value = 0.031, sample mean height is same as the population mean.

B.    p_value = 0.047, sample mean height is same as the population mean.

C.    p_value = 0.047, sample mean height is different from the population mean.

D.    p_value = 0.031, sample mean height is different from the population mean.

 

Correct Answer:p_value = 0.047, sample mean height is different from the population mean.

Explanation:

import numpy as np

from scipy import stats

'''

Null Hypothesis (H0): The sample mean height is equal to the population mean (μ = 175.3 cm).

Alternate Hypothesis (H1): The sample mean height is different from the population mean (μ ≠ 175.3 cm).

'''

# Population mean

population_mean = 175.3

# Heights of 10 UK men

survey_height = [177.3, 182.7, 169.6, 176.3, 180.3, 179.4, 178.5, 177.2, 181.8, 176.5]

# Perform one-sample t-test

t_statistic, p_value = stats.ttest_1samp(survey_height, population_mean,alternative = "two-sided")

# Set the significance level

alpha = 0.05

# Print the results

print("t-statistic:", t_statistic)

print("p-value:", p_value)

# Determine the conclusion

if p_value < alpha:

    print("Reject the null hypothesis")

    print("There is enough evidence to conclude that the sample mean height is different from the population mean.")

else:

    print("Fail to reject the null hypothesis")

    print("There is no enough evidence to conclude that the sample mean height is different from the population mean.")

Output:

t-statistic: 2.295568968083183
p-value: 0.04734137339747034
Reject the null hypothesis
There is enough evidence to conclude that the sample mean height is different from the population mean.

 

Q2. Two Schools IQ

Samples of IQ scores are collected from two competing schools, as follows:

school_1 = [115, 111, 112, 101, 95, 98, 100, 90, 89, 108]

school_2 = [107, 103, 91, 99, 104, 98, 117, 113, 92, 96, 108, 115, 116, 88]

Perform an appropriate test with a 5% significance level to check if there is any statistically significant difference in the mean IQ's of these schools.

Choose the correct option below :

A.    p_value = 0.56, IQ scores of the two schools are same

B.    p_value = 0.71, IQ scores of the two schools are same

C.    p_value = 0.56, IQ scores of the two schools are statistically different

D.    p_value = 0.71, IQ scores of the two schools are statistically different

Correct Option: p_value = 0.71, IQ of two schools are same

Explanation:
Based on the given problem, we define our hypothesis as:

• H0 : IQ scores of the two schools are same μ1​=μ2​
• Ha : IQ scores of the two schools are different μ1​​=μ2​

Since we need to compare two different samples, whose sample size is small (<30), we will use Two Sample T-test

Further, in order to test for the proposed alternate hypothesis, we will have to perform a Two Tailed Test.

Code:

from scipy.stats import ttest_ind

 

# Sample IQ scores

school_1 = [115, 111, 112, 101, 95, 98, 100, 90, 89, 108]

school_2 = [107, 103, 91, 99, 104, 98, 117, 113, 92, 96, 108, 115, 116, 88]

alpha = 0.05 # Significance level

 

# Performing Two sample Twp Tiled T-test

t_stat, pvalue = ttest_ind(school_1, school_2, alternative ='two-sided')

print(pvalue)

 

if pvalue < alpha:

  print('Reject H0 ; IQ scores of the two schools are different')

else:

  print('Fail to reject H0 ; IQ scores of the two schools are same')

Output:
0.7154458095186707
Fail to reject H0 ; IQ scores of the two schools are same

 

 

 

 

 

 

 

Q3. T/Z Test

The distributions of two variables x and y are given below in the form of boxplots.

Mark the correct option regarding variables x and y.

A.    We can use One sample t-test for variable x.

B.    We can use One sample t-test for variable y.

C.    We can't use One sample z-test for variable y.

D.    We can’t use both z-test or t-test for variable x.

E.     We can’t use both z-test or t-test for variable y.

 

Correct options:

• We can’t use both z-test or t-test for variable x.
• We can use One sample t-test for variable y.

Explanation:

Distribution of variable x:

• We know that the line inside the boxplot represents the mean.
• As clearly visible from the boxplot for x, it’s mean is not at the centre. It is significantly towards the left.
• When plotted, it’s peak will also shift towards left as follows which means it does not follow normal distribution.
• This is known as being highly skewed distribution (We will study skewness in detail in a few classes).
• Therefore, neither of Z or T-test can be used for distribution x.

Distribution of variable y:

• Here, the distribution of variable y is normal.
• Hence, we can use Z-test or T-test.

 

 

 

Q4. Select the Test

A pharmaceutical company is developing a new drug to decrease the blood pressure. In a clinical trial, they measured the blood pressure of 30 participants before administering the new drug.

After a month of treatment, they measured the blood pressure again for the same group. The company wants to determine if there is a significant decrease in blood pressure after using the new drug.

Which hypothesis test should the company use, and what are the characteristics of the test?

A.    One-tailed, Paired T-test

B.    One-tailed, One-sample, Z-test

C.    One-tailed, Two-sample, T-test

D.    Two-tailed, Paired T-test

E.     Two-tailed, Two-sample, T-test

Correct Option: One-tailed, Paired T-test

Explanation:
In this scenario, the company is interested in assessing whether there is a significant decrease in blood pressure after using the new drug.

• Since the blood pressure is measured for the same participants before and after treatment, it involves paired observations. Therefore, a Paired T-test is suitable.

Null and Alternative Hypotheses:

• Null hypothesis (H₀): There is no statistically significant difference in the average blood pressure before and after the new drug treatment.(µ_difference = 0), where µ_difference is the mean difference in blood pressure.
• Alternative hypothesis (H₁): There is a statistically significant decrease in the average blood pressure after the new drug treatment.(µ_difference < 0) assuming the drug aims to lower blood pressure.

The test is one-tailed because the focus is in a particular direction of the distribution.