jilliand2242
07.04.2020 •
Mathematics
A population is known to be normally distributed with a standard deviation of 2.8. (a) Compute the 95% confidence interval on the mean based on the following sample of nine: 8, 9, 10, 13, 14, 16, 17, 20, 21. (b) Now compute the 99% confidence interval using the same data.
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Ответ:
Step-by-step explanation:
Mean = sum of terms/number of terms
The mean of the given sample is
(8 + 9 + 10 + 13 + 14 + 16 + 17 + 20 + 21)/9 = 14.2
a)
For a confidence level of 95%, the corresponding z value is 1.96.
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
Where
n represents the number of samples
It becomes
14.2 ± 1.96 × 2.8/√9
= 14.2 ± 1.96 × 0.933
= 14.2 ± 1.83
The lower end of the confidence interval is 12.37
The upper end of the confidence interval is 16.03
b)
For a confidence level of 99%, the corresponding z value is 2.58
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
Where
n represents the number of samples
It becomes
14.2 ± 2.58 × 2.8/√9
= 14.2 ± 2.58 × 0.933
= 14.2 ± 2.4
The lower end of the confidence interval is 11.8
The upper end of the confidence interval is 16.6
Ответ:
90=9x
90/9=x
10=x
3x=3*10=30
6x=6*10=60
hope this helped