Question

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

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 ​(

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)

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