Easter candy expenditure per consumer is normally distributed with a standard deviation of $4.01. A candy manufacturer claims that the average Easter candy expenditure per consumer is no less than $30. Fifteen consumers were randomly selected. The average Easter candy expenditure per consumer was found to be $28.67 with a standard deviation of $3.50. Can you reject the candy manufacturer's claim at α=.05?
What is the 99% confidence interval for the true average Easter candy expenditure per consumer (in $)?
a. 
(26.6407, 30.6993) 

b. 
(26.0039, 31.3361) 

c. 
(26.9668, 30.3732) 

d. 
(25.9797, 31.3603) 

e. 
None of the answers is correct 
Solution :
Given that,
Point estimate = sample mean = = 28.67
sample standard deviation = s = 3.50
sample size = n = 15
Degrees of freedom = df = n  1 = 15  1 = 14
At 99% confidence level
= 1  99%
=1  0.99 =0.01
/2
= 0.005
t/2,df
= t0.005,14 = 2.977
Margin of error = E = t/2,df * (s /n)
= 2.977 * ( 3.50/ 15)
Margin of error = E = 2.6903
The 99% confidence interval estimate of the population mean is,
± E
= 28.67 ± 2.6903
= ( 25.9797, 31.3603 )
correct option is = d
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