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The decline of salmon fisheries along the Columbia River in Oregon has caused great concern among commercial and recreational fishermen. The paper 'Feeding of Predaceous Fishes on Out-Migrating Juvenile Salmonids in John Day Reservoir, Columbia River' (Trans. Amer. Fisheries Soc. (1991: 405-420) gave the accompanying data on 10 values for the data sets where y = maximum size of salmonids consumed by a northern squaw fish (the most abundant salmonid predator) and x = squawfish length, both in mm. Here is the computer software printout of the summary: Coefficients: Estimate Std. Error t value Pr(> |t|) (Intercept) −90.020 16.702 −5.390 0.000 Length 0.701 0.044 15.798 0.000 Using this information, compute a 95% confidence interval for the slope.

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nuhulawal20

Answer: { 0.5995, 0.8025 }

Step-by-step explanation:

Given that

                   Estimates          Std. Error         t value            Pr(>/t/)

Intercept:         -90.020          16.702           -5.390             0.000

length    :           0.701             0.044            15.798              0.000

Now using the given information to compute a 95% confidence interval for the slope:

We use the formula

β₁ ± tₐ/₂, ₙ₋₂ × ∝β₁

So we know that number of values (n) = 10

therefore error of degree of freedom df = n -2 = (10-2) = 8

Level of significance α ( 1 - 0.95 ) = 0.05

so tₐ/₂, ₙ₋₂ = t ₍₀.₀₅/₂, ₁₀₋₂

t ₀.₀₂₅, ₈ = 2.306 (critical value)

From the given table ( regression analysis output)

slope regression β₁ = 0.701

The standard error of the slope is Sβ₁  = 0.044

Let “the maximum size of salmonids consumed by a northern squaw fish” be the response variable and “squawfish length” be the explanatory variable.

The 95% confidence interval for the slope of the regression is:

β₁ ± tₐ/₂, ₙ₋₂ × ∝β₁ = 0.701 ± 2.306 (0.044)

= 0.701 ± 0.101464

= { 0.701 - 0.101464, 0.701 + 0.101464 }

= { 0.599536, 0.802464 } ≈  {0.5995, 0.8025 }

MrRoyal

The confidence interval of the slope is (0.599, 0.803)

The sample size is given as:

[tex]\mathbf{n = 10}[/tex]

The confidence interval is given as:

[tex]\mathbf{CI = 95\%}[/tex]

Start by calculating the degrees of freedom

[tex]\mathbf{df = n - 2}[/tex]

So, we have:

[tex]\mathbf{df = 10 - 2}[/tex]

[tex]\mathbf{df = 8}[/tex]

The level of significance is calculated as:

[tex]\mathbf{\alpha = 1 - CI}[/tex]

So, we have:

[tex]\mathbf{\alpha = 1 - 95\%}[/tex]

[tex]\mathbf{\alpha = 0.05}[/tex]

The critical value at 0.05 level of significance and 8 degrees of freedom is:

[tex]\mathbf{t_{\alpha} =2.306}[/tex]

The confidence interval of the slope is then calculated as:

[tex]\mathbf{CI = \beta_1 \pm t_\alpha \times S\beta_1}[/tex]

From the question, we have:

[tex]\mathbf{S\beta_1 = 0.044}[/tex] --- standard error of the slope

[tex]\mathbf{\beta_1 = 0.701}[/tex] -- the slope

So, the equation becomes

[tex]\mathbf{CI = \beta_1 \pm t_\alpha \times S\beta_1}[/tex]

[tex]\mathbf{CI = 0.701 \pm 2.306 \times 0.044}[/tex]

[tex]\mathbf{CI = 0.701 \pm 0.102}[/tex]

Split

[tex]\mathbf{CI = (0.701 - 0.102,0.701 + 0.102)}[/tex]

[tex]\mathbf{CI = (0.599,0.803)}[/tex]

Hence, the confidence interval of the slope is (0.599, 0.803)

Read more about confidence intervals at:

https://brainly.com/question/24131141

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