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 )

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!