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Sagot :
Answer:
Using a 95% confidence level, the lower end point of a two-tailed interval that would be used to predict the range of values that this 17th sampled unit would be in with respect to seconds soaking a component part in solution is of 29.137 seconds.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.1315
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.1315\frac{8}{\sqrt{16}} = 4.263[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 33.4 - 4.263 = 29.137 seconds.
Using a 95% confidence level, the lower end point of a two-tailed interval that would be used to predict the range of values that this 17th sampled unit would be in with respect to seconds soaking a component part in solution is of 29.137 seconds.
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