Question

Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared to standard gene fragments that can identify the​ species, 56​% were mislabeled. ​a) Construct a 90​% confidence interval for the proportion of all seafood sold in the country that is mislabeled or misidentified. ​b) Explain what your confidence interval says about seafood sold in the country. ​c) A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

Answer 1

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:

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

Verification:

$560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10$

Both the conditions are satisfied. This, the criticism is invalid.