Question

Question #a2ec7

Answer 1

Sample Mean `=35.5`
Sample Standard Deviation `~~8.155`
Given a 5% level of significance, these results have statistical significance.

Explanation:

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)/16`
`=568/16`

`=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=1.5` 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`
`=(1.5)^2`
`=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,`
`(27-35.5)^2=72.25 and (26-35.5)^2=90.25,`
`(39-35.5)^2=12.25 and(35-35.5)^2=0.25,`
`(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,`
`(21-35.5)^2=210.25 and (51-35.5)^2=240.25,`
`(28-35.5)^2=56.25 and (40-35.5)^2=20.25,`
`(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)/16`
`=1064/16`

`=66.5`

The last step is to take the square result of the previous result, called the variance , so:
`sqrt66.5~~8.155` is the standard deviation.

I hope I helped!