Therefore the solution of this problem is a)critic value t = 2.8982 b)margin of error= 2.985146
In statistics, the standard deviation measures how widely distributed or how much a set of values might vary. In contrast to a high standard deviation, which indicates that values are widely dispersed, a low standard deviation indicates that values are typically close to the set's mean.
Here ,
where s is the sample's standard deviation
Finding the number of degrees of freedom is the first step in solving this problem. The sample size has been reduced by 1.
So
df = 18 - 1 = 17
99% confidence interval
We must now determine the value of T, which can be done by consulting the t table, which has 17 degrees of freedom (on the y-axis) and a confidence level of).
[tex]1 - \frac{1-0.99}{2} =0.995[/tex]
So we have T = 2.8982.
The margin of error is:
M = T*s = 2.8982*1.03 = 2.985146
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