Question

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

Answer 1

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