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Provide an Appropriate Response Is There Evidence at The α=0.01\alpha = 0.01

Question 19

Multiple Choice

Provide an appropriate response.
-A soda machine is set to fill bottles with 24 ounces of cola. As part of a quality control program, an inspector records the amount of fill over or under 24 ounces for one minute. His observations are shown below.
0.400.450.210.150.710.050.840.120.050.580.971.010.750.500.080.450.270.750.071.000.250.250.910.990.360.050.110.310.840.670.450.600.010.980.220.720.770.480.130.390.730.780.900.140.680.800.120.160.370.27\begin{array} { r r r r r r r r r r } - 0.40 & - 0.45 & - 0.21 & - 0.15 & - 0.71 & - 0.05 & - 0.84 & - 0.12 & - 0.05 & - 0.58 \\ - 0.97 & - 1.01 & - 0.75 & 0.50 & 0.08 & - 0.45 & - 0.27 & - 0.75 & - 0.07 & - 1.00 \\ 0.25 & 0.25 & 0.91 & 0.99 & 0.36 & 0.05 & 0.11 & 0.31 & 0.84 & 0.67 \\ 0.45 & 0.60 & 0.01 & 0.98 & 0.22 & 0.72 & 0.77 & 0.48 & 0.13 & 0.39 \\ - 0.73 & - 0.78 & - 0.90 & - 0.14 & - 0.68 & 0.80 & 0.12 & 0.16 & 0.37 & 0.27 \end{array}
Is there evidence at the α=0.01\alpha = 0.01 level of significance to support the hypothesis that the filling process is not random with respect to amounts over and under 24 ounces?


A) z=5.71z = - 5.71
Since z>z0.005=2.58| \mathrm { z } | > \mathrm { z } _ { 0 } .005 = 2.58 , we reject H0\mathrm { H } _ { 0 } . There is sufficient evidence to support the hypothesis that the filling process is not random.
B) z=6.28z = - 6.28
Since z>z0.005=2.58| \mathrm { z } | > \mathrm { z } 0.005 = 2.58 , we reject H0\mathrm { H } _ { 0 } . There is sufficient evidence to support the hypothesis that the filling process is not random.
C) z=1.25\mathrm { z } = - 1.25
Since z<z0.005=2.58| z | < z _ { 0.005 } = 2.58 , we do not reject H0\mathrm { H } _ { 0 } . There is not sufficient evidence to support the hypothesis that the filling process is not random.
D) z=0.67z = 0.67
Since z<z0.005=2.58| z | < z _ { 0.005 } = 2.58 , we do not reject H0H _ { 0 } . There is not sufficient evidence to support the hypothesis that the filling process is not random.

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