Concerned about the initial monitoring data, the environmental action group decides to continue to monitor the plant, and try to gather more evidence for their case. A random sample of twenty-eight hours is selected over a period of a week. The observations (gallons of wastewater discharged per hour) are 887 1129 1095 1167 1162 1051 1146 904 1006 1302 1131 1145 1051 1091 982 1187 1068 1217 1107 1175 1187 1143 1206 975 942 1164 1027 1172 The output of a statistical analysis software on this dataset is shown below Mean 1100.7 N 28 Reading the output, we find that • The sample size is n = • The sample mean is π = • The sample standard deviation is s = S • From this we can calculate the standard error to be SE = √n || Std. Dev. 100.1

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Concerned about the initial monitoring data, the environmental action group decides to continue to
monitor the plant, and try to gather more evidence for their case. A random sample of twenty-eight hours
is selected over a period of a week. The observations (gallons of wastewater discharged per hour) are
887 1129 1095 1167 1162 982 1187 1068 1217 1107
1175 1187 1143 1206 975
1051 1146
904 1006
1302
1131 1145 1051 1091 942 1164 1027 1172
The output of a statistical analysis software on this dataset is shown below
Mean
N
28
Reading the output, we find that
●
The sample size is n =
The sample mean is
The sample standard deviation is s =
• From this we can calculate the standard error to be SE =
=
1100.7
=
||
$7/00
Std. Dev.
100.1
Transcribed Image Text:Concerned about the initial monitoring data, the environmental action group decides to continue to monitor the plant, and try to gather more evidence for their case. A random sample of twenty-eight hours is selected over a period of a week. The observations (gallons of wastewater discharged per hour) are 887 1129 1095 1167 1162 982 1187 1068 1217 1107 1175 1187 1143 1206 975 1051 1146 904 1006 1302 1131 1145 1051 1091 942 1164 1027 1172 The output of a statistical analysis software on this dataset is shown below Mean N 28 Reading the output, we find that ● The sample size is n = The sample mean is The sample standard deviation is s = • From this we can calculate the standard error to be SE = = 1100.7 = || $7/00 Std. Dev. 100.1
Note that the observed sample mean is is greater than 1000 gallons per hour. This could mean that the
plant is discharging more wastewater than they promised, or the plant could be in compliance, and the
large numbers were due to sampling variability. To see if this is the case, we will test the hypothesis that
1000 against the alternative that μ ‡ 1000 at a significance level of a = = 0.05.
μl
=
x - 1000
SE
The degrees of freedom for the t-score is n - 1
=
• Calculate the t-score t =
=
If we look up the critical value for this problem in the t-table, we would find that it is 2.052.
• The magnitude of the t-score is (O smaller greater) than the critical value. Therefore we should (
O reject fail to reject) the hypothesis that u = 1000
Therefore the evidence ( does not does) provide significant reason to believe that the plant is
discharging more wastewater on average than they claim to be.
Transcribed Image Text:Note that the observed sample mean is is greater than 1000 gallons per hour. This could mean that the plant is discharging more wastewater than they promised, or the plant could be in compliance, and the large numbers were due to sampling variability. To see if this is the case, we will test the hypothesis that 1000 against the alternative that μ ‡ 1000 at a significance level of a = = 0.05. μl = x - 1000 SE The degrees of freedom for the t-score is n - 1 = • Calculate the t-score t = = If we look up the critical value for this problem in the t-table, we would find that it is 2.052. • The magnitude of the t-score is (O smaller greater) than the critical value. Therefore we should ( O reject fail to reject) the hypothesis that u = 1000 Therefore the evidence ( does not does) provide significant reason to believe that the plant is discharging more wastewater on average than they claim to be.
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