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andreyandreev [35.5K]
2 years ago
13

A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and

the sample standard deviation is s = 15. Use Appendix D to find the values of Student's t. (a) Construct an interval estimate of μ with 98% confidence. (Round your answers to 3 decimal places.) The 98% confidence interval is from to
Mathematics
1 answer:
gayaneshka [121]2 years ago
4 0

Answer:

(832.156, \ 847.844)

Step-by-step explanation:

Given data :

Sample standard deviation, s = 15

Sample mean, \overline x = 840

n = 23

a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}

$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$

$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$

∴ $E = 2.508 \times \frac{15}{\sqrt{23}}$

  $E = 7.844$

So, 98% CI is

$(\overline x - E, \overline x + E)$

(840-7.844 ,  \ 840+7.844)

(832.156, \ 847.844)

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