The cost would be $582.40!
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Answer:
18/(10-7)+2=8
(48/2)-4^2 +2*2=12
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Answer:
The p-value of the test statistic from the standard normal table is 0.0017 which is less than the level of significance therefore, the null hypothesis would be rejected and it can be concluded that there is sufficient evidence to support the claim that less than 20% of the pumps are inaccurate.
Step-by-step explanation:
Here, 1304 gas pumps were not pumping accurately and 5689 pumps were accurate.
x = 1304, n = 1304 + 5689 = 6993
The level of significance = 0.01
The sample proportion of pump which is not pumping accurately can be calculated as,
The claim is that the industry representative less than 20% of the pumps are inaccurate.
The hypothesis can be constructed as:
H0: p = 0.20
H1: p < 0.20
The one-sample proportion Z test will be used.
The test statistic value can be obtained as:

Answer:
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Step-by-step explanation:
Sample size, n = 51
p = 0.62
1 - p = 1 - 0.62 = 0.38
n = 515
Confidence level = 90% = Zcritical at 90% = 1.645
Confidence interval = (p ± margin of error)
Margin of Error = Zcritical * sqrt[(p(1-p))/n]
Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]
Margin of Error = 1.645 * 0.0214
Margin of Error = 0.035203
Lower boundary = (0.62 - 0.035203) = 0.584797
Upper boundary = (0.62 + 0.035203) = 0.655203
(0.5848 ; 0.6552)
We are confident that about 58% to 66% of sea foods in the country are Mislabelled.
No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.
Answer:
It should be 62
Step-by-step explanation:
30+40-8=62