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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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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500

Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

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Collective rate of both, $\frac{d(m+w)}{dt}$ = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min = 20 min

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⇒ $w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft$

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate $\frac{dh}{dt}$ .

Lets calculate h.

Applying Pythagoras formula.

⇒ $h^2=(m+w)^2+500^2$

⇒ $h=\sqrt{(2400+2700)^2+500^2} = 5124.45$

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ $h^2=(m+w)^2+500^2$

⇒ $2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}$

⇒ $\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}$ ...as $\frac{d(500)}{dt}= 0$

⇒ Plugging the values.

⇒ $\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}$ ...as $\frac{d(m+w)}{dt} = 5$ ft/sec

⇒ $\frac{d(h)}{dt} =4.9761$ ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

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