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Anastasy [175]
3 years ago
8

The Mamestra brassicae a moth is primarily known as a pest that is responsible for severe crop damage on a wide variety of plant

species. Because there are limited means of chemically controlling the species, the use of wasp species that parasitize the eggs is used to control the reproduction and hence presence of these moths in agricultural crops. At a farm where a wasp species was introduced the eggs of the moths where collected and examined to determine if the wasps had successfully parasitized the eggs of the moths. Of 100 eggs collected 64 were determined to have been taken over by the wasps. Determine the 95% confidence interval for the true proportion moth eggs parasitized by the wasps.
Engineering
1 answer:
expeople1 [14]3 years ago
3 0

The 95% confidence interval for the true proportion moth eggs parasitized by the wasps is;

CI = (0.5459, 0.7341)

We are given;

Sample size; n = 100

Number of successes in the sample; x = 64

Now, formula for estimated proportion of success is; p = x/n

p = 64/100

p = 0.64

Formula for confidence interval of proportions is;

CI = p ± z√(p(1 - p)/n)

Where z is the critical value at the confidence level.

From tables, at 95% CL, critical value z = 1.96

Thus;

CI = 0.64 ± 1.96√(0.64(1 - 0.64)/100)

CI = 0.64 ± 0.0941

Thus;

CI = (0.64 - 0.0941), (0.64 + 0.0941)

CI = (0.5459, 0.7341)

Read more about confidence interval at;brainly.com/question/17042789

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The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

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