Answer to Question #191675 in Statistics and Probability for Josh Lucky Roblo

Question

A company which produces batteries claims that the life expectancy of their batteries is 90 hours. In order to test the claim, a consumer interest group tested a random sample of 40 batteries. The test resulted to a mean lif expectancy of 87 hours. Usoing a 0.05 level of significance, can it concluded that the life expectancy of their batteries is less than 90 hours? Assume that the population standard deviation is known to be 10 hours.

Answer 1

$H_0:$ The mean life expectancy of their batteries is 90 hours, i.e. $H_0: μ= 90$

$H_1:$ The mean life expectancy of their batteries is less than 90 hours, i.e. $H_1: μ < 90$

n=40 (large sample)

$\bar{x}=87 hours$

σ=10 hours

μ= 90

Level of significance$= α =0.05$

Since the sample size is large we can use the test statistic as Z

Test statistics: 

$Z = \dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

            $= \dfrac{87-90}{ \frac{10}{\sqrt{ 40}}}$

           $= \dfrac{-3}{ 1.58}$

           $= - 1.90$

P-value$= P(Z < - 1.90)= P(Z > 1.90)$

                   $=0.5 – 0.4713$ {Value taken from standard normal table}

                   =0.0287

   0.0287 < 0.05

   P-value < level of significance

  Reject $H_0$ .

Therefore, we have sufficient evidence to conclude that the mean life expectancy of their batteries is less than 90 hours .