Question

A sample of 17 observations is selected from a normal population. The sample standard deviation is 21.75, and the sample mean is 36.

a. Determine the standard error of the mean. (Round the final answer to 4 decimal places.)

The standard error of the mean is            .

b. Determine the 80% confidence interval for the population mean. (Round the final answers to 2 decimal places.)

   

The 80% confidence interval for the population mean is between  and  .

   

c. f you wanted a wider interval, would you increase or decrease the confidence level?

(Click to select)  Increase  Decrease

Answer 1

Solution :

Given that,

= 36

s =21.75

n =17

Degrees of freedom = df = n - 1 = - 17 = 16

a ) At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

  / 2= 0.20 / 2 = 0.10

t _/2,df = t_0.10,16 =1.337 ( using student t table)

standard error =(s /n) =(21.75 / 17) =5.2751

Margin of error = E = t_/2,df * (s /n)

= 1.337 * (21.75 / 17)

E=7.05

The 80% confidence interval estimate of the population mean is,

- E < < + E

36 - 7.05 < <36 + 7.05

28.95 < < 43.05

( 28.95 and  43.05)