To find the mean , we add all the values together, then divide them by how many numbers there were, so:
(37+43+30+27+26+39+35+38+45+46+42+21+51+28+40+20)/1637+43+30+27+26+39+35+38+45+46+42+21+51+28+40+2016
=568/16=56816
=35.5=35.5, therefore the average age of the customers from this sample is 35.5 years old.
With a 5% level of significance, any luck or chance based events should affect the score only within
30*0.05=1.530⋅0.05=1.5 points. Since the average from this sample is higher then the level of significance, we can say that the score has statistical significance , and isn't merely due to a sampling error, thus rejecting the null hypothesis .
To find the standard deviation, we take the square root of the average squared distances from the mean. It looks like this:
(37-35.5)^2(37−35.5)2
=(1.5)^2=(1.5)2
=2.25=2.25. We do this for the rest of the values to get:
(43-35.5)^2=56.25 and (30-35.5)^2=30.25,(43−35.5)2=56.25and(30−35.5)2=30.25,
(27-35.5)^2=72.25 and (26-35.5)^2=90.25,(27−35.5)2=72.25and(26−35.5)2=90.25,
(39-35.5)^2=12.25 and(35-35.5)^2=0.25,(39−35.5)2=12.25and(35−35.5)2=0.25,
(38-35.5)^2=6.25, and(45-35.5)^2=90.25,(38−35.5)2=6.25,and(45−35.5)2=90.25,
(46-35.5)^2=110.25 and (42-35.5)^2=42.25,(46−35.5)2=110.25and(42−35.5)2=42.25,
(21-35.5)^2=210.25 and (51-35.5)^2=240.25,(21−35.5)2=210.25and(51−35.5)2=240.25,
(28-35.5)^2=56.25 and (40-35.5)^2=20.25,(28−35.5)2=56.25and(40−35.5)2=20.25,
(20-35.5)^2=240.25(20−35.5)2=240.25.
Now that we have all the values, we add them together then divide them by how many there are, so
(2.25+56.25+30.25+72.25+90.25+12.25+0.25+6.25+90.25+110.25+42.25+210.25+240.25+56.25+20.25+240.25)/162.25+56.25+30.25+72.25+90.25+12.25+0.25+6.25+90.25+110.25+42.25+210.25+240.25+56.25+20.25+240.2516
=1064/16=106416
=66.5=66.5
The last step is to take the square result of the previous result, called the variance , so:
sqrt66.5~~8.155√66.5≈8.155 is the standard deviation.
I hope I helped!
