Help with math homework

Question

Use the following sample to estimate a population mean .

74.6

65.9

76.1

57.3

65.9

74.4

Assuming the population is normally distributed, find the 99.5% confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places.

99.5% C.I. = 

Answer 1

Hi Morgan,

For any confidence interval, you will need a sample mean (xbar) and standard deviation (s). Computing standard deviation by hand is cumbersome and error-prone, so use a graphing calculator (TI-80s series). From TI-83 Plus:

xbar= 69.03

s= 7.31

Now, we were not given a population standard deviation, which means we need to use a t-confidence interval. Formula for that is:

CI = xbar +/- t*SE

xbar=sample mean

t*=t-critical value obtained from t-table or software

SE=standard error

SE= s/sqrt(n)

For our problem:

xbar= 69.03

s= 7.31

n= 6

SE= 7.31/sqrt(6)

SE=2.98

Now, we need t8, which means we also need degrees of freedom:

df= n-1

df= 6-1 = 5

Now, navigate to a t-table and look up 99.5% confidence in the column and 5 degrees of freedom in the row:

t* = 4.773

Now we have everything we need to compute our confidence interval:

CI = xbar +/- t*SE

xbar= 69.03

t* = 4.773

SE = 2.98

CI= 69.03 +/- (4.773*2.98)

CI = (54.81, 83.25)

I hope this helps.