Answer:
The calculated value t = 2.083 < 2.429 at 0.01 level of significance with 38 degrees of freedom.
The null hypothesis is accepted
That is x⁻ and 'μ' do not differ significantly.
Step-by-step explanation:
<u>Step(i)</u>:-
<em>Given sample size 'n' = 39 </em>
<em>The mean of the Population 'μ' = 8</em>
<em>The mean of the sample x⁻ = 8.2</em>
The standard deviation of the sample 'S' = 0.6
Level of significance ∝=0.01
<u>Step(ii)</u>:-
<em>Null hypothesis : H₀ : x⁻ = 'μ' </em>
<em>Alternative hypothesis: H₁: x⁻ ≠ 'μ' </em>
<em>Degrees of freedom ν = n-1 = 39-1 =38</em>
<em> t₀.₉₉ = 2.429 ( see t- table)</em>
The test statistic
t = 2.083 < 2.429 at 0.01 level of significance
<u>Step (iii)</u>:-
<em>The calculated value t = 2.083 < 2.429 at 0.01 level of significance with 38 degrees of freedom.</em>
<em>The null hypothesis is accepted </em>
<em>That is x⁻ and 'μ' do not differ significantly.</em>