Please show work Suppose a grocery store is considering the purchase of a new se

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Please show work

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 h faster t he 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 Sample Sample standard deviation (min) Standard error estimate (min) SE, = 3.97143 SE-= 3.27321 size mean (min) =126.4 =111.0 new machine nl=49 31 = 27.8 old machine "-: 46 22.2 s dr 90.71233 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 new machine minus the old machine. so that a positive result reflects faster checkour times with the new machine, Use the Satterthwaite approximate degrees of freedom, 90.71233. Give your answers precise to at least three decimal places lower = upper =

Explanation / Answer

here we use non-pooled t-statistic to calculate the confidence interval since the given df=90.71

SE(difference of means)=sqrt(s12/n1+s22/n2)=sqrt(27.8*27.8/49+22.2*22.2/46)=5.15

(1-alpha)*100% confidence interval for difference of means=difference ±t(alpha/2,df)*SE(difference)

95% confidence interval for diffeence of population mean=(126.4-111)±t(0.05/2, 90.71)*5.15

=15.4±1.99*5.15=15.4±10.25=(5.15, 25.65)

(second part) meaning of lower and upper limit of confidence interval

right choice is second.

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