Question

7. In a sample of 100 boxes of a certain type, the average compressive strength was 6230 N, and the standard a. Find a 95% confidence interval for the mean com- b Find a 99% confidence interval for the mean com- c. An engineer claims that the mean strength is be- deviation was 221 N pressive strength of boxes of this type. pressive strength of boxes of this type. tween 6205 and 6255 N. With what level of confi dence can this statement be made? d. Approximately how many boxes must be sampled so that a 95% confidence interval will specify the mean to within ±25 N?

Answer 1

Here we have : n = 100, $\overline{x}$ = 6230, s = 221

a. The 95% confidence interval is given by

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

Where,

$E=Zc *\frac{s}{\sqrt{n}}$

c = 0.95 , $\alpha =1-c=1-0.95 = 0.05$

$Zc=Z_{1-\alpha /2}=Z_{1-0.05/2}=1.96$

So,

$E=1.96*\frac{221}{\sqrt{100}}$

= 43.32

Hence the confidence interval is given by

( 6230 - 43.32 , 6230 + 43.32 )

( 6186.68, 6273.32 )

b. The 99% confidence interval is given by,

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

Where,

$E=Zc *\frac{s}{\sqrt{n}}$

c = 0.99 , $\alpha =1-c=1-0.99 = 0.01$

$Zc=Z_{1-\alpha /2}=Z_{1-0.01/2}=2.58$

So,

$E=2.58*\frac{221}{\sqrt{100}}$

= 57.018

Hence the confidence interval is given by

( 6230 - 57.018 , 6230 + 57.018 )

( 6172.982, 6287.018 )

C. Here the confidence interval is ( 6205, 6255 )

The confidence interval formula is

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

So ,  $\overline{x}$ - E = 6205

  $\overline{x}$ + E = 6255

-----------------------------

$2*\overline{x}$ = 12460

$\overline{x}$ = 6230

so ,     $\overline{x}$ + E = 6255

6230 + E = 6255

E = 25

$Zc*\frac{s}{\sqrt{n}}=25$

$Zc*\frac{221}{\sqrt{100}}=25$

Zc * 22.1 = 25

Zc = $Z_{1-\frac{\alpha }{2}}$ = 1.13

Using normal probability table , we get probability 0.871

$1-\frac{\alpha }{2}=0.87$

$1-0.87=\frac{\alpha }{2}$

$\alpha =0.26$

c = 1 - $\alpha$ = 1 - 0.26 = 0.74

Hence the level of confidence is 0.74.

d. Here we have c = 0.95 , E = 25 , Zc = 1.96

$n = \left ( \frac{Zc* s }{E} \right )^{2}$

  $= \left ( \frac{1.96* 221 }{25} \right )^{2}$

= 300.20

$\approx$ 301

Hence the required sample size is 301.