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).

d. Construct a 90​% confidence interval estimate. How does this change your answer to part​ (b)?

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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 0.996

sample standard deviation = s = 0.01

sample size = n = 50

Degrees of freedom = df = n - 1 = 49

d)

At 90% confidence level the t is ,

t _/2,df = t_0.05,49 = 1.677

Margin of error = E = t_/2,df * (s /n)

= 1.677 * ( 0.01/ 50)

= 0.00237

The 90% confidence interval estimate of the population mean is,

- E < < + E

0.996 - 0.00237 < < 0.996 + 0.00237

0.99363 < < 0.99837