Question

The manager of a paint supply store wants to estimate the actual amount of paint contained in 1​-gallon cans purchased from a nationally known manufacturer. The​ manufacturer's specifications state that the standard deviation of the amount of paint is equal to 0.01 gallon. A random sample of 50 cans is​ selected, and the sample mean amount of paint per 1​-gallon can is 0.996 gallon. Complete parts​ (a) through​ (d).

a. Construct a 95​% confidence interval estimate for the population mean amount of paint included in a​ 1-gallon can. a. Construct a 95​% confidence interval estimate for the population mean amount of paint included in a​ 1-gallon can.

nothingless than or equalsmuless than or equals nothing ​(Round to five decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 0.996

= 0.01

n = 50

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (0.01 / 50)

= 0.00277

At 95% confidence interval estimate of the population mean is,

- E < < + E

0.996 - 0.00277 < < 0.996 + 0.00277

0.99323< < 99877

(0.99323 ,  99877 )