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.)
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 = t0.05,49 = 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
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