The manager of a paint supply store wants to determine wheth

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 0.02 gallon. you select a random sample of 50 cans and the mean amount of paint per 1-gallon can is 0.995 gallon. a, is there evidence that the mean amount is different from 1.0 gallon(use =0..01)? b, compute the p-value and interpret its meaning. c, Construct a 99% confidence interval estimate of the population mean amount of paint. d, Compare the result of (a) and (c) . what conclusions do you reach?

Solution

A)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   1  
Ha:    u   =/   1  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical z, as alpha =    0.01   ,      
alpha/2 =    0.005          
zcrit =    +/-   2.575829304      
              
Getting the test statistic, as              
              
X = sample mean =    0.995          
uo = hypothesized mean =    1          
n = sample size =    50          
s = standard deviation =    0.02          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.767766953          
              
Also, the p value is              
              
p =    0.077099872          
              
Comparing |z| < 2.576, (or, p > 0.01), we   FAIL TO REJECT THE NULL HYPOTHESIS.          

Thus, there is no significant evidence that the mean amount is different from 1.0 gallon. [CONCLUSION]

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b)

As we saw,

p =    0.077099872          

Thus, the probability of getting at least a value at least this extremely far from the mean is p =    0.077099872.

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c)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    0.995          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    0.02          
n = sample size =    50          
              
Thus,              
Margin of Error E =    0.007285545          
Lower bound =    0.987714455          
Upper bound =    1.002285545          
              
Thus, the confidence interval is              
              
(   0.987714455   ,   1.002285545   ) [ANSWER]

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d)

The conclusion is the same. In part C, we still fail to reject Ho since 1.00 is inside the interval.