Test the significant for large sample maths mini project

TEST THE SIGNIFICANCE FOR
LARGE SAMPLE
1. Brajesh Bhaskar Kodam ARMIET/CS23/KB015
2. Manish Choudhary ARMIET/AIML23/SA004
3. Rameshwar Jadhav ARMIET/AIML23/VA016
4. Vishal Yadav ARMIET/AIML23/AT011
(Branch): Comp
Semester-IV

INTRODUCTION
In statistics, we often deal with data that varies—like people’s
weight, height, or other measurable values. To make sense of this
data, we usually take a small part of it, called a sample, and
compare it with what we expect from the entire group, or
population. This helps us:
 Estimate important values about the whole group
(population).
 Study key features of the population based on a sample.
When the sample size is more than 30, it’s usually considered a
large sample. The techniques we use for testing significance in
large samples are a bit different from those used for small
samples. These methods help us draw more accurate and
trustworthy conclusions.

THEORY
Significance testing is a statistical method that helps us figure out
whether the results we see in a sample are just due to random
chance, or if they actually reflect something true about the whole
population. It plays a key role in making smart, data-based
decisions by testing different assumptions (called hypotheses).
In general, when a sample has more than 30 observations (n >
30), it's considered a large sample. Large sample tests are very
common in research and analysis because:
 They provide more reliable results than small sample
tests.
 The normal approximation applies, making statistical
calculations easier.
 They help in making decisions based on data, such as
comparing group means or proportions.

COMMON TESTS FOR LARGE SAMPLES
1. Z-Test (For Large Samples)
 One-Sample Z-Test: Comparing a sample mean to a known
population mean.
 Two-Sample Z-Test: Comparing means of two independent
samples.
 Proportion Z-Test: Testing a sample proportion against a
claimed proportion.
2. Chi-Square Test (For Categorical Data)
 Independence Test: It is used to check whether two
categorical variables are related or independent. (It
answers: "Is there a relationship between two factors?")
 Goodness-of-Fit Test: It is a statistical test used to see if
observed data matches expected data based on a specific
theoretical distribution.(It answer: "Does this data follow
the pattern we expected?")

STEPS IN SIGNIFICANCE TESTING FOR
LARGE SAMPLES
1. State the Null (H₀) and Alternative Hypothesis (Ha)
2. Compute the Test Statistic
 Use the formula based on the test type.
3. Choose a Significance Level ()
 Common values: 0.05 (5%) or 0.01 (1%).
4. Compare with Critical Value or P-Value
 If the test statistic exceeds the critical value or p-value <
α, reject H .
₀
5. Make a Conclusion
 If H is rejected, it means there is a significant effect or
₀
difference.
 If H is not rejected, the difference is likely due to
₀
chance.

PROBLEM STATEMENT: (One Sample Z test)
Q1. A tyre company claims that the lives of tyres have a mean of 42,000 km
with a standard deviation of 4,000 km. A new product is tested on a sample
of 81 tyres, and the sample mean is found to be 42,500 km. Test at 5% level
of significance whether the new product is significantly better than the old
one.
Solution:
n= 81 tyres, = 42500 , µ= 42000, = 4000
Step 1: Null Hypothesis(H₀):The mean tyre life of the new product is not
significantly different from the old product. (µ= 42000)
Alternative Hypothesis (Ha): The mean tyre life of the new
product
is significantly different from the old product. (µ 42000)
Step 2: Test Statistic:
Z= = = 1.125

PROBLEM STATEMENT: (One Sample Z test)
Step 3: Level of Significance()
= 5% i.e. 0.05
Step 4: Critical Value (Z)
1. To find this refer the z-table.
2. Since it is a two tailed test 0.05/2 = 0.025
3. Critical value is 1-0.025 = 0.9750
4. 0.9750 = 0.5000 + value in table
Hence, Z = 1.96
Step 5: Z < Z
Hence, we fail to reject H .
₀
This means the new tyre is not significantly
better than the old one at the 5% significance
level.
Two tailed test
Z table

PROBLEM STATEMENT: (Two Sample Z test)
Q2. A researcher wants to compare test scores between two schools.
•School A: n = 50, mean = 78, SD = 10
•School B: n = 60, mean = 75, SD = 12
At a 5% significance level, is there a significant difference in the average
scores?
Solution:
•Step 1: Null Hypothesis (H )
₀ : There is no difference in average scores. (μ₁
= μ )
₂
Alternative Hypothesis (H )
ₐ : There is a significant difference
in average scores. (μ ≠ μ )
₁ ₂
Z= = = 1.43

PROBLEM STATEMENT: (Two Sample Z test)
= 5% i.e. 0.05
4. 0.9750 = 0.5000 + value in table
Hence, Z = 1.96
Step 5: Z < Z
₀
There is not enough evidence to conclude that the average test
scores of the two schools are significantly different.
Z table

PROBLEM STATEMENT: (Proportion-Z Test)
Q3. A mobile company claims that 60% of customers are satisfied
with their service.
A survey of 200 customers finds that 120 are satisfied.
Test the company’s claim at a 5% significance level.
Solution:
Step 1: Null Hypothesis(H₀): The true proportion of satisfied
customers is 60%.(p= 0.60)
Alternative Hypothesis (Ha): The true proportion is not
60%. (p 0.60)
Z= == 0 …… =

PROBLEM STATEMENT: (Proportion-Z Test)
= 5% i.e. 0.05
4. 0.9750 = 0.5000 + value in table
Hence, Z = 1.96
Step 5: Z < Z
₀
There is not enough evidence to reject the company’s claim.
The proportion of satisfied customers is consistent with 60%.
Z table

PROBLEM STATEMENT: (Chi-square test)
Q4. A principal wants to determine if student absences are equally
distributed across the weekdays. A sample of 100 teachers reports the
highest absence days, with the observed and expected counts given
below. At a 5% significance level, do absences occur equally across all
days? Use the Chi-Square Goodness of Fit Test to determine the answer.
DAYS MONDAY TUESDAY WEDNESDAY THURSDAY FRIDAY
Observed 23 16 14 19 28
Expected 20 20 20 20 20
Solution:
Step1: Null Hypothesis(H₀):Student absence are equally distributed across
all weekdays.
Alternative Hypothesis(Ha):Student absence not equally distributed
across weekdays.

Step 2: Test Statistics:
= = Observed Value, = Expected value
DAYS OBSERVED EXPECTED /
MONDAY 23 20 3 9 0.45
TUESDAY 16 20 -4 16 0.8
WEDNESDAY 14 20 -6 36 1.8
THURSDAY 19 20 -1 1 0.05
FRIDAY 28 20 8 64 3.2
= 0.45+0.8+1.8+0.05+3.2= 6.3
= 5% i.e. 0.05

Step 4: Critical value:
df= 5-1 = 4
df= 4 = 9.488… From Chi- Square table.
Step 5: 6.3 < 9.488
Hence, we fail to reject the null hypothesis (H )
₀
There is no significant evidence to suggest that student absences
are not equally distributed across weekdays. The absences appear
to occur uniformly across the week.
Chi-square table

Q5. A marketing analyst wants to test if people use five different
social media platforms equally. Out of 150 people surveyed, the
observed usage distribution is as follows. Test at α = 0.05 to
determine if the usage is evenly distributed.
PLATFORMS INSTAGRAM FACEBOOK TWITTER SNAPCHAT LINKEDIN
Observed 30 25 35 40 20
Expected 30 30 30 30 30
Solution:
Step1: Null Hypothesis(H₀): Social media usage is evenly distributed across
platforms.
Alternative Hypothesis(Ha): Social media usage is not evenly
distributed.

Step 2: Test Statistics:
PLATFORMS OBSERVED EXPECTED /
INSTAGRAM 30 30 0 0 0
FACEBOOK 25 30 -5 25 0.833
TWITTER 35 30 5 25 0.833
SNAPCHAT 40 30 10 100 3.333
LINKEDIN 20 30 -10 100 3.333
= 0+0.833+0.833+3.333+3.333= 8.332
= 5% i.e. 0.05

df= 5-1 = 4
Step 5: 8.332 < 9.488
Hence, we fail to reject the null hypothesis (H )
₀
There is no significant evidence to suggest that social media usage
is unevenly distributed — the usage appears roughly equal across
platforms.
Chi-square table

Q6. A health researcher wants to determine whether a person’s diet type is
related to the presence of health issues. To investigate this, a survey was
conducted on 100 individuals, and the number of people with and without
health issues was recorded for each diet category. At a 5% level of
significance, can we conclude that diet type and health issues are
associated?
Diet type Has Health Issues No Health Issues Total
Vegetarian 10 30 40
Non-Vegetarian 25 15 40
Vegan 5 15 20
Total 40 60 100
Solution:
Step1: Null Hypothesis(H₀): Diet type and health issues are independent
Alternative Hypothesis(Ha): Diet type and health issues are not

Step 2: Expected frequencies
E=
Diet Type Has Issues(E) No Issues (E)
Vegetarian (40×40)/100 = 16 (60×40)/100 = 24
Non-Vegetarian (40×40)/100 = 16 (60×40)/100 = 24
Vegan (40×20)/100 = 8 (60×20)/100 = 12
Step 3:Test Statistics:
Observed Expected /
Vegetarian, Has Issues 10 16 -6 36 2.25
Vegetarian, No Issues 30 24 6 36 1.5
Non-Veg, Has Issues 25 16 9 81 5.06
Non-Veg, No Issues 15 24 -9 81 3.38
Vegan, Has Issues 5 8 -3 9 1.13
Vegan, No Issues 15 12 3 9 0.75
Total 14.07

= 5% i.e. 0.05
df= (r-1)(c-1)= (3-1)(2-1)= 2
Step 6: 14.07 > 5.991
Hence, we reject the null hypothesis (H )
₀
There is a significant relationship between diet type and health
issue.
Chi-square table

CONCLUSION
• Significance testing helps us check if a sample truly represents
the whole group (population). When the sample size is large
(more than 30), special methods are used to get accurate
results.
It helps us:
 Estimate important values about the whole group.
 Understand patterns and differences in data.
 Make better decisions based on numbers, not guesses.
• In short, significance testing in large samples helps us study
data correctly and make reliable conclusions.

APPLICATION
Significance testing for large samples is widely used in various fields, including:
1. Medical Research & Drug Testing:
 Used in clinical trials to determine if a new drug is more effective than an
existing treatment.
2. Quality Control in Manufacturing:
 Ensures products meet required standards by testing sample batches.
3. Market Research & Consumer Behavior:
 Companies test whether a new product’s average sales differ significantly
from a competitor.
4. Education & Psychological Studies:
 Used to compare student performance before and after an educational
intervention.

REFERENCES:
 Fisher, R. A. (1925). Statistical Methods for Research
Workers. Edinburgh: Oliver and Boyd.
 Neyman, J., & Pearson, E. S. (1933). On the Problem of
the Most Efficient Tests of Statistical Hypotheses.
Philosophical Transactions of the Royal Society of London.

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