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Sagot :
Answer:
a) The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is 10.02.
Step-by-step explanation:
The t-distribution is used to solve this question:
a) n = 14
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 = 14 - 1 = 13
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.7709
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 1.7709\frac{40}{\sqrt{14}} = 18.93[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) n = 28
27 df, T = 1.7033
[tex]M = T\frac{s}{\sqrt{n}} = 1.7033\frac{40}{\sqrt{28}} = 12.88[/tex]
The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is
44 df, T = 1.6802
[tex]M = T\frac{s}{\sqrt{n}} = 1.6802\frac{40}{\sqrt{45}} = 10.02[/tex]
The margin of error for a 90% confidence interval when n = 45 is 10.02.
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