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Sagot :
Answer:
The calculated value t = 1.399 < 2.5705 at 0.025 level of significance
Null hypothesis is accepted
The population distribution is approximately normal
Step-by-step explanation:
Step(i):-
Given mean of the Population = 5.0
sample size 'n'= 6
The mean of the sample x=5.4 pounds /square inch
The standard deviation of the sample (S) = 0.7
Level of significance = 0.025
Null Hypothesis : The sample came from population
H₀ : x⁻ = μ
Alternative Hypothesis :H₁ : x⁻ ≠ μ
Step(ii):-
Test statistic
[tex]t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{5.4 -5.0}{\frac{0.7}{\sqrt{6} } }[/tex]
t = 1.399
Degrees of freedom = n-1 =6-1 =5
The tabulated value
t = 2.5705
The calculated value t = 1.399 < 2.5705 at 0.025 level of significance
Null hypothesis is accepted
The population distribution is approximately normal
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