Question

A random sample of 16 pharmacy customers showed the waiting times below (in minutes).

26	16	25	18	17	24	18	23
14	20	10	18	19	13	17	16

Click here for the Excel Data File

Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population. (Round your standard deviation answer to 4 decimal places and t-value to 3 decimal places. Round your answers to 3 decimal places.)
  
The 90% confidence interval  to

Answer 1

Given sample data

24,16,25,18,17,24,18,23,14,20,10,18,19,13,17,16

Sample Mean = Sum of values / number of values

= 292 /16

= 18.25

Standard deviation S = (xi - )2 / n-1

where xi are the sample values of x , n =16, ​​​​​​​ = 18.25

(xi - ​​​​​​​ )2  = (24 - 18.25)2 + (16 - 18.25)2 +(25 - 18.25)2 +(18 - 18.25)2 +(17 - 18.25)2 +(24 - 18.25)2 +(18 - 18.25)2 +(23 - 18.25)2 +(14 - 18.25)2 +(20 - 18.25)2 +(10 - 18.25)2 +(18 - 18.25)2 +(19 - 18.25)2 +(13 - 18.25)2 +(17 - 18.25)2 + (16 - 18.25)2

(xi - ​​​​​​​ )2  = 265 , n-1 =15

Standard deviation S = 265 / 15

= 17.666

= 4.203173

Standard deviation = 4.2032 rounded to 4 decimals places

t-value will be obtaine dy the degrees of freedom and value

degrees of freedom = n-1 = 15

= 1 - confidence interval

= 1 - 0.9

= 0.1

The t-value for df=9 and = 0.1 will be 1.753 which can be calculated by the below attached table

90% confidence interval for the mean = t-value * S /

Where S is the standard deviation of the sample

90% confidence interval for the mean = 18.25 1.753 * 4.2032 /

= 18.25 1.753 * 4.2032 / 4

= 18.25 1.753 * 1.0508

= 18.25 1.842052

= (16.40795, 20.09205)

90% confidence interval for the mean is (16.40795, 20.09205)

Standard deviation = 4.2032

t-value = 1.753