Question

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 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 ((Intercept) ~89.010 16.720 [Length 0.704 0.049 t value 55324 14.333 Pr(> Itl) 0.000 0.000 Using this information , give the equation of the least squares regression line and the 95% confidence interval for the slope. a) y =16.720x 89.010; [0.623 , 0.785] b) y =0.704x 89.010; [0.604, 0.804] =0.704x + 16.720; [-121.781,-56.239] y =16.720x + 0.049; [0.349,-54.780] e y =-89.010x + 0.704; [-122.985, -55.035] f) None of the above

Answer 1

The 95% confidence interval for the slope is (0.594, 0.816)

From the table, we have the following parameters

b = 0.704 --- the slope

sb = 0.049 --- the standard error of the slope

n = 10 -- the sample size

The degree of freedom is calculated using

df = n = 2

So, we have:

df = 10 - 2

df = 8

At 95% confidence interval, and degrees of freedom of 8;

The critical value is:

$t _\frac{\alpha}{2} = 2.31$

The confidence interval is then calculated as:

CI = (b+/-Sb * tα/2)

This gives

CI = (0.704 +/- 0.049 * 2.31)

CI = (0.704 +/- 0.113)

Split

CI = (0.704 - 0.113, 0.704 + 0.113)

CI = (0.591, 0.817)

Hence the answer is, the 95% confidence interval is (0.591, 0.817).

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