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. 1425 1227 989 697 729 839 729 742 550 626 1449 941 ​(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 95​% confidence interval for the population mean mu. a 95​% confidence interval for the population mean is ​( nothing​, nothing​). ​(round to one decimal place as​ needed.)

Answer 1

A) $\overline{x}=911.9$
B) s = 304.2
C) $911.9\pm172.1$

To find the sample mean, add all of the data points together and then divide by 12, the number of data points:
(1425+1227+989+697+729+839+729+742+550+626+1449+941)/12 = 10943/12 = 911.9

To find the sample standard deviation, subtract each data point from the mean; square the result; find the average of these; then take the square root.

To construct a confidence interval, first find the z-score that goes along with it:
Convert 95% to a decimal:  0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value corresponds with a z-score of 1.96.

We use this to calculate the margin of error:
$ME=z*\frac{s}{\sqrt{n}} \\ \\=1.96*\frac{304.2}{\sqrt{12}}=172.1$

This gives us the confidence interval
$\overline{x}\pm ME \\ \\911.9\pm 172.1$