1) The document discusses the t-test for a single mean, which is used to test hypotheses about population means using small samples (n ≤ 30) that are assumed to be normally distributed. 2) It provides the formula for the t-test statistic and explains how to compare it to critical t-values to reject or fail to reject the null hypothesis. 3) An example applies the t-test to test if the average thickness of a sample of 10 washers matches the expected average thickness. The null hypothesis is not rejected.

1. DEPARTMENT OF MBA
A presentation on
TEST OF SIGNIFICANCE OF T-TEST FOR SINGLE MEAN WITH PROBLEM
Submitted by
19L31E0007 MALLA JAYA CHANDRA

2. SMALL SAMPLE THEORY
• Generally the number of observations in a sample alre equal or less than 30. than that sample is
said to be small sample. These small samples are based on two fundamental assumptions.
• The parent population is which the sample are assumed to follow a normal distribution.
• The sample observations are independent.
• To test the significance of population parameters of small sample. We apply exact (t,f, X2 )
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3. DIFFERENT TYPES OF SMALL SAMPLES TESTS
• Small sample tests for single mean
• Small sample test for difference of mean.
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4. SMALL SAMPLE TEST FOR SINGLE MEAN (T-
TEST FOR SINGLE MEAN)
Let (𝑥1, 𝑥2, … . . 𝑥𝑛) Are the observations of a sample of size n (𝑛 ≤ 30). Drawn from a normal
population of mean µ and variance 𝜎2
The test weather the population mean µ.is equal to again specified to µ0 the null hypothesis is given
by
H0: µ=µ0
AGAINST
H1 It depends on the question.
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5. Under H0 THE TEST STATISTIC FOR T TEST FOR SINGLE MEAN IS GIVEN BY
𝑡 =
𝑥 − 𝑢
𝑆 ∕ 𝑛
∞𝑡 𝑛−1
(If data is given directly)
𝑡 =
𝜘 − 𝜇
𝑠 ∕ ℎ − 1
~𝑡 𝑛−1
If individual data is given
Now obtain the calculated value of t and compare with tabulated value of t for degree of freedom
(n-1) at a given level of significance.
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6. • If 𝑡𝑐𝑎𝑙 𝑜𝑟 𝑡𝑐𝑎𝑙 ≤Ttab then the H0 may be accepted.
• If 𝑡𝑐𝑎𝑙 𝑜𝑟 𝑡𝑐𝑎𝑙 >Ttab then the H0 may be rejected.
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7. LEVEL OF SIGNIFICANCE
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Level of significance One tailed (H1)(<,>) Two tailed(≠)
1% 0.05 0.05/2=0.025
5% 0.01 0.01/2=0.005

8. EXAMPLE PROBLEM
• A machine is designed to produce insulated washers for electronic devices of average 0.025 cms.
A random sample of 10 washers was found to have an average thickness of 0.024 cms. With a
standard deviation of 0.002 cms. Test the significance of mean at 0.05.
• H0 µ = 0.025
• Against
• H1 µ≠ 0.025
• N=10, µ=0.025, 𝑥 = 0.024, S=0.002
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9. 19L31E0007 MALLA JAYA CHANDRA 9
• 𝑡 =
𝑥−𝑢
𝑆∕ 𝑛
∞𝑡 𝑛−1
• Substituting all the values in the above formula
• )/𝑡 = (0.024 − 0.025)/(0.002)/ 10
• Tcal=-1.6667
• |tcal|=1.6667
• T tab for(10-1=9) at level of significance 5% for two tailed is 2.262.

10. • Therefore |tcal|<t tab
• H0 is accepted.
• µ = 0.025
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11. THANKING YOU FOR GIVING THIS
WONDERFUL OPPORTUNITY
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