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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

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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