Question

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. 1420 1227 989 691 726 833 722 747 547 630 1442 948 ​(a) Find the sample mean. x overbarequals nothing ​(Round to one decimal place as​ needed.) ​(b) Find the sample standard deviation. sequals nothing ​(Round to one decimal place as​ needed.) ​(c) Construct a 90​% confidence interval for the population mean mu. A 90​% confidence interval for the population mean is ​(

Answer 1

n = 12

Data : 1420, 1227, 989, 691, 726, 833, 722, 747, 547, 630, 1442, 948

(a) Mean = sum of terms / no of terms = 10922 / 12 = 910.1667

Mean = 910.2

(b) standard deviation = s

data	data-mean	(data - mean)2
1420	509.8333	259929.99378889
1227	316.8333	100383.33998889
989	78.8333	6214.68918889
691	-219.1667	48034.04238889
726	-184.1667	33917.37338889
833	-77.1667	5954.69958889
722	-188.1667	35406.70698889
747	-163.1667	26623.37198889
547	-363.1667	131890.05198889
630	-280.1667	78493.37978889
1442	531.8333	282846.65898889
948	37.8333	1431.35858889

Standard deviation = 303. 2

(c) Confidence interval

t value for 90% of confidence interval at 11 df is TINV(0.10,11) = 1.796

CI = mean +/- E = (753.013 , 1067.387)

CI = (753.0 , 1067.4)