Mean Length

Question

How can this be calculated?

 

After a particular wet night ,12 worms surfaced on the lawn. Their lengths measured in cm, were
9.5, 9.5 , 11.2 , 10.6 , 9.9 , 11.1 , 10.9 , 9.8 , 10.1 , 10.2 , 10.9 , 11.0
Assuming that this sample comes from a normal population with variance 4 , calculate a 95% confidence interval for the mean length of all the warms in the garden.

Answer 1

We will use the following

 

sample average ± margin of error

 

 

 

We start by finding the sample average of the 12 worms:  (add all 12 values and divide by 12)

 

Now we will find the margin of error.

 

Since the variance of population is 4 the standard deviation is 2 since variance is the square of standard deviation.

 

Margin of Error (MOE):    Z* sigma/sqrt(n)

 

Z* is based on the level of confidence . We can look this up in a z-chart.  

 

We look up .975 in z-chart (.95 + 1tail(.025))

 

The value for 95% confidence is 1.96

 

Our sigma is 2 , the standard deviation. n = 12 (12 worms)

 

So our MOE = 1.96(2/sqrt(12)) = 1.1316

 

Now, add 1.1316 to the sample average (I will let you calculate the average of the 12 worms)

 

This would be your upper value of your interval

 

Also, subtract 1.1316 from the average (This is the lower value of the interval)

 

The interval will always be 2 margin of errors long (1 to the left and 1 to the right of the mean) and the center will always be the sample mean.

 

If you need more help, please let me know