Question

A company manufacturing stereo equipment claims that their personal CD player can be used for approximately 8 hours of continuous play when used with alkaline batteries. To provide this estimate, the company tested 35 CD players with new alkaline batteries and recorded the time at which the batteries in the players “lost power”. The average time was 8.3 hours with a sample standard deviation of 1.2 hours.

1. Construct a 98% confidence interval for the mean time until a new alkaline battery used in the CD player loses power.

CI = _____________________________

2. Determine the required sample size in order to estimate the mean time until a new alkaline battery used in the CD player loses power to within ±10 minutes at the 99% confidence level.

______________________________

Answer 1

Solution

a)

The 98% confidence interval for population mean is given as follows :

Where, x̅ is sample mean, s is sample standard deviation, n is sample size and t_{(0.02/2, n - 1)} is critical t value to construct 98% confidence interval.

We have,  x̅ = 8.3 hours, s = 72 minutes = 1.2 hours, n = 35

Using t-table we get, t_{(0.02/2, 35 - 1)} = 2.441

Hence, 98% confidence interval for the mean time untill a new alkaline battery used in the CD player loses power is,

= (8.3 ± 0.495)

= (8.3 - 0.495, 8.3 + 0.495)

= (7.805, 8.795)

The 98% confidence interval for the mean time untill a new alkaline battery used in the CD player loses power is (7.805 hours, 8.975 hours).

b)

for E = 10, 99% Confidence interval for Z = 2.576, s = 1.2hours = 72 min

sample size n = (Z*s/E)^2 = (2.576*72/10)^2 = 343.99 = 344