Question

989 The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. Assume the population is normally distributed. 1424 695 725 623 1221 721 745 1442 837 544 947 (a) Find the sample mean. X= (Round to one decimal place as needed.) (b) Find the sample standard deviation. s= (Round to one decimal place as needed.) (c) Construct a 95% confidence interval for the population mean H. A 95% confidence interval for the population mean is (1.0). (Round to one decimal nlace as needed)

Answer 1

Solution:

x	x2
1424	2027776
695	483025
725	525625
623	388129
1221	1490841
721	519841
745	555025
1442	2079364
989	978121
837	700569
544	295936
947	896809
$\sum$x=10913	$\sum$x2=10941061

The sample mean is $\bar x$

Mean $\bar x$ = ($\sum$x / n) )

=1424+695+725+623+1221+721+745+1442+989+837+544+947 /12

=10913/12

=909.4167

Mean $\bar x$ = 909.4

The sample standard is S

  S =$\sqrt{}$( $\sum$ x2 ) - (( $\sum$ x)2 / n ) n -1

=$\sqrt{}$10941061-(10913)212/11

=$\sqrt{}$10941061-9924464.0833/11

=$\sqrt{}$1016596.9167/11

=$\sqrt{}$92417.90/15

=304.0031

The sample standard is =304.0

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 95% confidence level the t is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,11 =2.201

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.201 * (304.0 / $\sqrt$ 12)

= 193.1

Margin of error = 0.50

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

909.4- 193.1 < $\mu$ < 909.4 + 193.1

716.2 < $\mu$ < 1102.5

(716.2, 1102.5)