Answer:
a) Test statistic
Z = 1.265 < 1.96 at 0.05 level of significance
The battery life is not exceeds 40 hours
b)
p- value = 0.8962
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given sample size 'n' =10
Mean of the sample x⁻ = 40.5 hours
Mean of of the Population μ = 40 hours
Standard deviation of the Population = 1.25 hours
<u><em>Step(ii):-</em></u>
<u><em>Null Hypothesis:H₀</em></u>: μ = 40 hours
<u><em>Alternative Hypothesis :</em></u>H₁ : μ < 40 hours
<u><em>step(ii):-</em></u>
<em>Test statistic </em>
![Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bx%5E%7B-%7D%20-mean%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D)
![Z = \frac{40.5 -40}{\frac{1.25}{\sqrt{10} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B40.5%20-40%7D%7B%5Cfrac%7B1.25%7D%7B%5Csqrt%7B10%7D%20%7D%20%7D)
Z = 1.265
Level of significance = 0.05
Z₀.₀₅ = 1.96
Z = 1.265 < 1.96 at 0.05 level of significance
The battery life is not exceeds 40 hours
<u><em>Step(iii):</em></u>-
<u><em>P - value </em></u>
P( Z < 1.265) = 0.5 + A( 1.265)
= 0.5 + 0.3962
= 0.8962
P( Z < 1.265) = 0.8962
i ) p- value = 0.8962 > 0.05
Accept H₀
There is no significant
The battery life is not exceeds 40 hours