1 answer

989 The state test scores for 12 randomly selected high school seniors are shown on the...

Question:

989 The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c)

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)

Answers

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
\sumx=10913 \sumx2=10941061

The sample mean is \bar x

Mean \bar x = (\sumx / 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 /\sqrtn)

= 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)

.

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