Please write 300-400 words Sup

Steps:

1. First of all, we specify our variable under study. Here it isx or ‘life of the battery’. Therefore the assumed mean, orµ₀=500 and σ=25.
2. At this point, we need to collect a sample of battery liveslarge enough to test for significance of the difference between theactual and the assumed means. Hence, we collect a large sample,i.e. the lives in days of 30 or more batteries. For ease ofcalculation in this case, let n be equal to 30.
3. Here, H₀: µ=µ₀=500 and H₁:µ<µ₀=500
4. Say this sample gives a mean of 480. If this number was 150,183 or 310, we could have rejected the null hypothesis right away.However, since 480 looks close enough to 500, we test thisstatistically using the z-statistic.
5. z= (µ-µ₀)/(σ/sqrt(n)) = (480-500)/(25/(30^0.5)) =-4.38 which is more than two standard deviations away from themean.
6. Testing this in R, using pnorm(480,500,25/sqrt(30)) we get avalue smaller than 0.01, which will be ourp-value. This means that, when the population meanis 500, there is less than 1% probability that we would find asample mean smaller than 480.
7. To see whether the above probability is small enough, we checkat a particular level of significance. This depends upon the usageof the batteries. Say they are to be used in pacemakers – in thiscase, the consequences of a type II error could be fatal (ie. thebattery life is taken to be 500 days). Hence, we take α to be0.1.
8. Now, the p-value (~0.01) is less than α (0.1), therefore, wemight want to advise the end consumers to switch batteries with asafe margin in hand, such as once every year if the sample mean isaround 480.