Question

Trap spacing measurements (in meters) for a sample of seven teams of red spiny lobster fisherman are reproduced in the accompanying table. Let u represent the average of the trap spacing measurements for the population of red spiny lobster fisherman. The mean and the standard deviation of the sample measurements are x = 90.3 meters and s = 12.1 meters, respectively. Suppose you want to determine if the true value of u differs from 95 meters. Complete parts a through h below. 93 99 107 95 82 70 86

Answer 1

X 93 X - mean 2.7143 8.7143 16.7143 4.7143 -8.2857 -20.2857 -4.2857 (X-mean 2 7.3674 75.939 279.368 22.2246 68.6528 411.51 18.3672


Since we know that

Mean() - Σ1 ;


Where n is the number of data points
Now

Στη = 632.

and n = 7

This implies that

Mean(ī) = 0.2.0 = 90.2857

Since we know that

Variance(s?) = n -1 ti - 7)2 = 883.4284 n=7 883.4284 Variance(s) = Variance(s?) = 147.2381 Standard Deviation(s) = V Variance Standard Deviation(s) = 12.1342 6

FOR HYPOTHESIS

No = 95 n=7 s= 12.1 X = 90.3 a = 0.05


The test hypothesis is

Null Hypothesis - - > Hou = Ho Alternate Hypothesis - -> H: Huo

This is a two-sided test because the alternative hypothesis is formulated to detect differences from the hypothesized mean value of 30 on either side.

Now, the value of test static can be found out by following formula:

X - LLO to = s/ n 90.3 - 95.0 12.1/17 to = -1.0277 to


Using Excel's function
=T.DIST.2T (tol,n - 1)
,
the P-value for
to = -1.0277
in an t-test with 6 degrees of freedom can be computed as
P = P(T. > -1.0277) = T.DIST.27(1 – 1.02771, 6) = 0.3437

Since, P = 0.34372634416616366 > 0.05, we fail to reject the null hypothesis
Hou = 95.0 at a = 0.05


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