Question

In: Statistics and Probability

The Maryland Fisheries Service releases an annual Oyster Population Status Report that includes an oyster biomass...

The Maryland Fisheries Service releases an annual Oyster Population Status Report that includes an oyster biomass index, catch landings data, and mortality rates. During the 2019-2020 season, the researchers obtained an oyster sample from 43 beds out of approximately 1000 oyster beds to assess the health of the stock.

Marine researchers want to be reasonably certain that the true proportion of Chesapeake Bay oysters caught at market size is at least 30 percent, the minimum threshold for qualifying the stock as “restored.” They found that the average proportion of market-size oysters in their sample was 0.39.

a) Calculate a 90 percent confidence interval for the population proportion of market-size oysters in the Chesapeake Bay in the 2019-20 season. Use a picture of the probability distribution in your answer.

b) Calculate a 99 percent confidence interval for the proportion of market-size oysters in the Chesapeake Bay in the 2019-20season. Use a picture of the probability distribution in your answer

c) Put in words what the 99% confidence interval tells us.

d) Conduct a hypothesis test with a significance of α=0.05 to determine if researchers can reasonably classify the Chesapeake Bay oyster stock

Solutions

Expert Solution

Answer:

Given that,

The Maryland Fisheries Service releases an annual Oyster Population Status Report that includes an oyster biomass index, catch landings data, and mortality rates.

During the 2019-2020 season, the researchers obtained an oyster sample from 43 beds out of approximately 1000 oyster beds to assess the health of the stock.

Marine researchers want to be reasonably certain that the true proportion of Chesapeake Bay oysters caught at market size is at least 30 percent, the minimum threshold for qualifying the stock as “restored.”

(a).

Note that this is a one-tailed test. In other words, the lower value of the confidence interval has no significance level and we only concerned with the upper bound (Test is for H_1: p < 0.3).

=NORMSINV(0.90)=1.28

=0.30

n=43

Sp=SQRT((1-)/n)=SQRT(0.39 (1-0.39)/43)=0.074

Upper bound of the C.I=0.39+1.28 0.074=0.485

So, the 90% C.I should be written as (-inf, 0.485).

Note:

If one intends to find the 2-tailed C.I then should be used C.I will be [0.39-1.645 0.074, 0.39-1.645 0.074]

i.e, [0.268, 0.512]

(b).

For 99%,

Upper bound of the C.I=0.39+2.326 ​​​​​​​ 0.076=0.56

So, the 99% C.I should be written as (-inf, 0.56).

Note:

C.I(2-tailed)=[0.39-2.576 0.39-1.645 ​​​​​​​ 0.074, 0.39+1.645 ​​​​​​​ 0.074]

i.e,[0.199, 0.581]

(c).

The confidence interval tells us that there is a 99% probability that the observed mean in a given sample drawn from the population will be within this interval.

(d).

Construct the 95% confidence interval and check whether the target mean proportion p0=0.3 lies in that interval.

=NORMINV(0.95)=1.645

=0.39

n=43

Sp=SQRT((1-)/n)

=SQRT(0.39(1-0.39)/43)=0.074

Upper bound of the C.I=0.39+1.645 0.074

=0.512

So, the 95% C.I should be written as (inf, 0.512).

Our target mean p0=0.3 is within this interval. So, we cannot reject the claim of H0. So, the true proportion is not less than 0.30.

They found that the average proportion of market-size oysters in their sample was 0.39.


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