Question

Suppose a grocery store is considering the purchase of a new self-checkout machine that will get customers through the checkout line faster than their current machine. Before he spends the money on the equipment, he wants to know how much faster the customers will check out compared to the current machine. The store manager recorded the checkout times, in seconds, for a randomly selected sample of checkouts from each machine. The summary statistics are provided in the table. Group Description Sample size Sample mean (min) Sample standard deviation (min) Standard error estimate (min) 1 old machine ?1=46 ?⎯⎯⎯1=123.4 ?1=25.8 SE1=3.80400 2 new machine ?2=43 ?⎯⎯⎯2=108.0 ?2=20.2 SE2=3.08047 df=84.45717 Compute the lower and upper limits of a 95% confidence interval to estimate the difference of the mean checkout times for all customers. Estimate the difference for the old machine minus the new machine, so that a positive result reflects faster checkout times with the new machine. Use the Satterthwaite approximate degrees of freedom, 84.45717. Give your answers precise to at least three decimal places.

Answer 1

Suppose a grocery store is considering the purchase of a new self-checkout machine that will get customers through the checkout line faster than their current machine. Before he spends the money on the equipment, he wants to know how much faster the customers will check out compared to the current machine. The store manager recorded the checkout times, in seconds, for a randomly selected sample of checkouts from each machine. The summary statistics are provided in the table. Group Description Sample size Sample mean (min) Sample standard deviation (min) Standard error estimate (min)

1 old machine ?1=46 ?⎯⎯⎯1=123.4 ?1=25.8 SE1=3.80400

2 new machine ?2=43 ?⎯⎯⎯2=108.0 ?2=20.2 SE2=3.08047 df=84.45717

Compute the lower and upper limits of a 95% confidence interval to estimate the difference of the mean checkout times for all customers. Estimate the difference for the old machine minus the new machine, so that a positive result reflects faster checkout times with the new machine. Use the Satterthwaite approximate degrees of freedom, 84.45717. Give your answers precise to at least three decimal places.

Table value t at 84 DF is 1.9886

95% CI for mean difference =(5.6660, 25.1340)

	old	new
	123.4	108	mean
	25.8	20.2	std. dev.
	46	43	n
		84	df
		15.4000	difference (old - new)
		4.8949	standard error of difference
		5.6660	confidence interval 95.% lower
		25.1340	confidence interval 95.% upper
		9.7340	margin of error