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Answer:
<em>The calculated value t = 2.038< 2.145 at 0.05 level of significance</em>
<em>Null hypothesis is accepted </em>
<em>There is the average rate is less than μ ≤ 4</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>The Population of the mean 'μ' =4</em>
<em>sample size 'n' = 16</em>
<em>sample mean 'x⁻' = 4.53</em>
<em>given sample standard deviation 's' = 1.04</em>
<em>level of significance α = 0.05</em>
<u><em>Step(ii)</em></u><em>:-</em>
<u><em>Null hypothesis:H₀</em></u> : There is no significance difference between two means
<u>Alternative hypothesis : H₁</u><em>: There is significance difference between two means</em>
Test statistic
<em> </em><em />
<em> </em>
t = 2.038
<em>Degrees of freedom ν = n-1 = 16-1 =15</em>
t₀.₀₂₅ = 2.145
<u><em>Conclusion</em></u>:-
<em>The calculated value t = 2.038< 2.145 at 0.05 level of significance</em>
<em>Null hypothesis is accepted </em>
<em>There is the average rate is less than μ ≤ 4</em>