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2 years ago

Helppp meeeeeeeeeeeeeeeeeee

Mathematics

2 answers:

dsp732 years ago

7 0

Answer:

-2/-3

Step-by-step explanation:

Just do it

Veronika [31]2 years ago

5 0

Answer:

First

Step-by-step explanation:

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Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

frutty [35]

Answer:

a) (0.5256,0.5944)

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

$\hat{p} = 0.56$

a) 90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

Putting the values, we get:

$0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)$

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

$np > 10\\n(1-p)>10$