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.
3/5 = 72/x
First u need to cross multiply
5*72 = 360
Now divide
360/3 =120. <<< that's x
So
3/5 = 72/120 < answer
Answer:
7x + 168.25
Step-by-step explanation:
Umm I wasn't completly sure but good luck!
