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We are interested in comparing the average supermarket prices of two leading colas in the Tampa area. Our sample was taken by randomly going to each of eight supermarkets and recording the price of a six-pack of cola of each brand. The data are shown in the following table. Find a 98% confidence interval for the difference in mean price of brand 1 and brand 2. Assume that the paired data came from a population that is normally distributed. Price Supermarket Brand 1 Brand 2 $2.25 $2.30 2.47 2.45 2.44 2 4 5 6 7 8 2.38 2.27 2.15 2.25 2.36 2.37 x1 = 2.3125 $1 = 0.1007 2.29 2.25 2.25 2.42 2.40 x2=2.3500 $2=0.0859 (-0.1768, 0.1018) (-0.0846, 0.0096) (-0.0722, -0.0028) (-0.0779, 0.0029) Difference $-0.05 0.02 -0.06 -0.02 -0.10 0.00 -0.06 -0.03 d=-0.0375 Sd = 0.0381

We are interested in comparing the average supermarket prices of two leading colas in the Tampa area. Our sample was taken by randomly going to each of eight supermarkets and recording the price of a six-pack of cola of each brand. The data are shown in the following table. Find a 98% confidence interval for the difference in mean price of brand 1 and brand 2. Assume that the paired data came from a population that is normally distributed. Price Supermarket Brand 1 Brand 2 $2.25 $2.30 2.47 2.45 2.44 2 4 5 6 7 8 2.38 2.27 2.15 2.25 2.36 2.37 x1 = 2.3125 $1 = 0.1007 2.29 2.25 2.25 2.42 2.40 x2=2.3500 $2=0.0859 (-0.1768, 0.1018) (-0.0846, 0.0096) (-0.0722, -0.0028) (-0.0779, 0.0029) Difference $-0.05 0.02 -0.06 -0.02 -0.10 0.00 -0.06 -0.03 d=-0.0375 Sd = 0.0381

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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We are interested in comparing the average supermarket prices of two leading colas in the Tampa area. Our sample was taken by randomly going to each of eight supermarkets and recording the price of a six-pack of cola of each brand. The data are shown in the following table. Find a 98% confidence interval for the difference in mean price of brand 1 and brand 2. Assume that the paired data came from a population that is normally distributed. Price Supermarket 1 2 3 4 5 8 Brand 1 $2.25 2.47 2.38 2.27 2.15 2.25 2.36 2.37 x1 = 2.3125 $1 = 0.1007 Brand 2 $2.30 2.45 2.44 2.29 2.25 2.25 2.42 2.40 x2 = 2.3500 $2 = 0.0859 (-0.1768, 0.1018) (-0.0846, 0.0096) (-0.0722,-0.0028) (-0.0779, 0.0029) Difference $-0.05 0.02 -0.06 -0.02 -0.10 0.00 -0.06 -0.03 d = -0.0375 sd = 0.0381
Transcribed Image Text:We are interested in comparing the average supermarket prices of two leading colas in the Tampa area. Our sample was taken by randomly going to each of eight supermarkets and recording the price of a six-pack of cola of each brand. The data are shown in the following table. Find a 98% confidence interval for the difference in mean price of brand 1 and brand 2. Assume that the paired data came from a population that is normally distributed. Price Supermarket 1 2 3 4 5 8 Brand 1 $2.25 2.47 2.38 2.27 2.15 2.25 2.36 2.37 x1 = 2.3125 $1 = 0.1007 Brand 2 $2.30 2.45 2.44 2.29 2.25 2.25 2.42 2.40 x2 = 2.3500 $2 = 0.0859 (-0.1768, 0.1018) (-0.0846, 0.0096) (-0.0722,-0.0028) (-0.0779, 0.0029) Difference $-0.05 0.02 -0.06 -0.02 -0.10 0.00 -0.06 -0.03 d = -0.0375 sd = 0.0381
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