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.

{ 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 }

yor answer is 7/6 times 5

step-by-step explanation:

51-50

step-by-step explanation: