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The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer is actually 1 gallon. You know from the manufacturer's specifications that the standard deviation of the amount of paint is 0.011 gallon. You select a random sample of 45 cans, and the mean amount of paint per 1-gallon can is 0.995 gallon. Complete parts (a) through (e) below. 


a. Is there evidence that the mean amount is different from 1.0 gallon? (Use a = 0.05.)

Let u be the population mean. Determine the null hypothesis, H0, and the alternative hypothesis, H1.

What is the test statistic? 


What is/are the critical value(s)? (Use a=0.05.)
 

First, start your hypothesis testing by stating the null and the alternative hypotheses. 

Null hypothesis,

H0: mu = 0.995
Ha: mu != 0.995

The standard normal z-score should be used because this is a normally distributed data. 

z= (sample mean - hypothesized mean)/(known standard deviation / sqrt(n))

z= (0.995 - 1)/(0.011/sqrt(45)) = -3.05

The Critical value at the 5% level of significance is 1.96. 

For a left-tailed test, the null hypothesis is rejected whenever the critical value is less than the test statistic. Therefore, the null hypothesis is rejected in this case.
There is enough evidence to suport the claim suggested. 

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