Suppose a grocery store is considering the purchase of a new self-checkout machine that will get...

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	n1=51n1=51	¯¯¯x1=128.4x¯1=128.4	s1=29.8s1=29.8	SE1=4.17283SE1=4.17283
2	new machine	n2=48n2=48	¯¯¯x2=113.0x¯2=113.0	s2=24.2s2=24.2	SE2=3.49297SE2=3.49297

df=94.99897df=94.99897

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, 94.99897. Give your answers precise to at least three decimal places.

upper limit:

lower limit:

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Expert Solution

orchestra answered 11 months ago

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