Question

Question 36 4 pts Use the confidence level and sample data to find a confidence interval for estimating the population P. Round your answer to the same number of decimal places as the sample mean. Test scores n-72 x=581, 0 = 6.2, 98% confidence 56.2 <<60.0 0 56.4 < p < 59.8 56.9< < 59.3 56.7 <<59.5

Answer 1

Solution :

Given,

$url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596056146&ymreqid=69d777f6-ff7f-1f97-1c77-9d008d01eb00&sig=y_sAy8IJJmgHd_U1MW9vhg--~D$ = 58.1   

$\sigma$ = 6.2

n = 72

Note that, Population standard deviation($\sigma$) is known..So we use z distribution. Our aim is to construct 98% confidence interval.

$\therefore$ c = 0.98

$\therefore$$\alpha$ = 1- c = 1- 0.98 = 0.02

$\therefore$  $\alpha$/2 = 0.02 $\slash$ 2 = 0.01 and 1- $\alpha$ /2 = 0.99

Search the probability 0.995 in the Z table and see corresponding z value

$\therefore$ = 2.326

2012

The margin of error is given by

E =  $url=https://latex.codecogs.com/gif.latex?Z&t=1596056146&ymreqid=69d777f6-ff7f-1f97-1c77-9d008d01eb00&sig=JUk1tqmbDcpwg2kg2AJTaQ--~D$$\alpha$/2 * ($\sigma$ / $\sqrt{}$ n )

= 2.326 * ( 6.2/ $\sqrt{}$ 72 )

= 1.700

Now , confidence interval for mean($\mu$) is given by:

($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596056146&ymreqid=69d777f6-ff7f-1f97-1c77-9d008d01eb00&sig=y_sAy8IJJmgHd_U1MW9vhg--~D$ - E ) <  $\mu$ <  ($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596056146&ymreqid=69d777f6-ff7f-1f97-1c77-9d008d01eb00&sig=y_sAy8IJJmgHd_U1MW9vhg--~D$ + E)

( 58.1 - 1.700 )   <  $\mu$ <  ( 58.1 + 1.700)

56.4 <  $\mu$ < 59.8