A marine biologist wanted to construct a t interval to estimate the mean weight of marine otters using 98% confidence. They took a random sample of n = 8 marine otters to measure their weights. These weights were roughly symmetric with a mean of 2 = 4.5 kg and a standard deviation of sx = 1.1 kg. What critical value t* should they use?

CI 98 %  =  ( 3.332  ;  5.668 )

t(c)  =  2.9979

Step-by-step explanation:

Confidence Interval  CI  98 %   then  α  =  2 %   α  =  0,02    α/2  =  0.01

Sample information:

Sample size    n  =  8

sample mean     x  =  4.5

standard deviation of sample    s  =  1.1 kg

degree of freedom     df  =  n  -  1     df  =  8  -  1    df  =  7

With    α/2  0.01  and  df  =  7 from t-studente table we find   t(c)

t(c)  =  2.9979

CI 98 %  =  (   x  ±  t(c) * s/√n )

CI 98 %  =  (  4.5  ±  2.9979 *  1.1/ √8 )

CI 98 %  =  (  4.5  ±  3.2976/2.8228 )

CI 98 %  =  (  4.5  ± 1.168 )

CI 98 %  =  ( 3.332  ;  5.668 )

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step-by-step explanation:

12

step-by-step explanation:

a triangle's   full angle is 180 degrees so 180 - 156 = 24. 24/2=12