Question

Many people think it’s faster to order at the drive-thru than to order inside at fast-food restaurants.  To find out, two students randomly selected 10 times over a 2-week period to visit a local Dunkin’ Donuts restaurant.  At each of these times, one student ordered an iced coffee at the drive-thru and the other student ordered an iced coffee at the counter inside.  The table shows the times, in seconds, that it took for each student to receive their iced coffee after the order was placed.

1. After calculating the difference (inside - outside) data, find the mean difference. Label as dbar =
2. Find the standard deviation of the difference data. Label as SD(d) =
3. State the inference procedure.
4. Check the conditions.
5. What critical number will you use for a 95% confidence interval.
6. Construct a 95% confidence interval. Use the format dbar +/- MOE = (xxx, xxx)
7. Interpret the confidence interval.
8. How many trips to DD’s are needed to estimate the true mean difference to within 2 seconds of the true value? Use your standard deviation from #2.

The key is to obtain the correct mean and standard deviation. Double and Triple check your differences. USE TECHNOLOGY (EXCEL or your GRAPHING CALCULATOR) TO OBTAIN THE MEAN AND STANDARD DEVIATION.

Visit	Inside	Outside
1	62	55
2	63	50
3	327	321
4	105	111
5	135	124
6	57	54
7	98	90
8	75	69
9	205	200
10	100	103

(Please type and answer all the questions to the full extent, thank you so much!)

Answer 1

Visit	Inside	Outside	differece =inside -outside
1	62	55	7
2	63	50	13
3	327	321	6
4	105	111	-6
5	135	124	11
6	57	54	3
7	98	90	8
8	75	69	6
9	205	200	5
10	100	103	-3

using minitab>stat>basic stat>paired t

we have

Paired T-Test and CI: Inside, Outside

Paired T for Inside - Outside

N Mean StDev SE Mean
Inside 10 122.7 84.3 26.7
Outside 10 117.7 84.4 26.7
Difference 10 5.00 5.81 1.84

95% CI for mean difference: (0.84, 9.16)
T-Test of mean difference = 0 (vs ≠ 0): T-Value = 2.72 P-Value = 0.024

mean difference ()= 5

the standard deviation of the difference data. SD(d) = 5.81

t at 95% with 9 degree of freedom = 2.262

the critical numberwe use for a 95% confidence interval is 2.262

95% confidence interval. = +/- MOE = +/-

=

=

=  0.84, 9.16

we are 95% confident that the mean difference in time that it took for each student to receive their iced coffee (0.84,9.16).

now for margin of error E = 2 sec

population standard deviation = 5.81

and z at 95% is 1.96

approximately 33  trips to DD’s are needed to estimate the true mean difference to within 2 seconds of the true value