Question

Duracell released an advanced long lasting AAA battery, which it claims can last on average for at least 200 days with the standard deviation of 16 days in any remote control devices. A random sample of 100 batteries was selected and the results showed that new batteries lasted on average for 198 days. Construct 90% confidence interval for the population mean and interpret it. Also conduct the five-step hypothesis test at α = 0.05 to determine whether the battery manufacturer’s claim is true or not and afterwards, compute the p-value for it.

Answer 1

Answer :

given data :-

(a) Level of Significance , α = 0.1

population std dev , σ = 16

Sample Size , n = 100

Sample Mean, x̅ = 198

z value= z α/2= 1.6449 [excel formula =normsinv(0.10/2) ]
Standard Error , SE = σ/√n

= 16/√100

= 16/10

= 1.6000

margin of error , E = Z*SE

= 1.6449 * 1.6000

= 2.6318

confidence interval is
Interval Lower Limit = x̅ - E

= 198-2.6318

= 195.3682

Interval Upper Limit = x̅ + E

= 198+2.6318

= 200.6318

we are 90% confident that population mean days for battery will lie within confidence interval.

(b) Ho : µ ≥ 200 (claim)

Ha : µ < 200

Level of Significance , α = 0.05
population std dev , σ = 16
Sample Size , n = 100
Sample Mean, x̅ = 198


Standard Error , SE = σ/√n

= 16/√100

= 16/10

= 1.6000

Z-test statistic= (x̅ - µ )/SE

= (198-200)/1.6000

= -2/1.6000

= -1.2500


p-Value = 0.1056 [excel formula , =normsdist(-1.25)

Conclusion: p-value>α, Do not reject null hypothesis

so, there isn't sufficient proof to dismiss the battery maker's case