Question

A company that manufactures electronic components wants to know the mean discharge time of one particular type of capacitor that it makes. In order to estimate this value, they randomly select 100 capacitors and measure how long they take to discharge. The mean of this sample was 7.25 seconds with a standard deviation of 0.15 seconds. Assuming that this distribution is normal, construct a 99% confidence interval to estimate the mean discharge time for all capacitors of this type. Since this is a Confidence Interval for a mean, what distribution would be used to calculate the critical value? Standard Normal Distribution t-Distribution Binomial Distribution You get to choose the distribution you prefer

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 7.25

Population standard deviation =    = 0.15

Sample size = n =100

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (0.15 / 100)

= 0.0386

At 99% confidence interval estimate of the population mean is,

- E < < + E

7.25 -0.0386 < < 7.25 +0.0386

7.2114< < 7.2886

(7.2114, 7.2886)