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Sagot :
Answer:
The best estimate for the length of gribbles is of 3.1 mm.
The margin of error for this estimate, with 95% confidence, is of 0.2 mm.
The 95% confidence interval is between 2.9 mm and 3.3 mm.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
Best estimate for the length of gribbles
The best estimate is the sample mean, that is 3.1 mm.
Margin of error:
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 = 50 - 1 = 49
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.01
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.01\frac{0.72}{\sqrt{50}} = 0.2[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error for this estimate, with 95% confidence, is of 0.2 mm.
Confidence interval:
The lower end of the interval is the sample mean subtracted by M. So it is 3.1 mm - 0.2 mm = 2.9 mm
The upper end of the interval is the sample mean added to M. So it is 3.1 mm + 0.2 mm = 3.3 mm
The 95% confidence interval is between 2.9 mm and 3.3 mm.
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