Question

Find the confidence interval specified. Assume that the population is normally distributed.

The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
8.3 7.3 13.8 12.6 11.1
9.5 13.7 14.2 9.8 12.6

Determine a 95% confidence interval for the mean time for all players

12.90 to 9.70 min

9.70 to 12.90 min

13.00 to 9.60

9.60 to 13.00 min

Answer 1

Solution:

x	x2
8.3	68.89
7.3	53.29
13.8	190.44
12.6	158.76
11.1	123.21
9.5	90.25
13.7	187.69
14.2	201.64
9.8	96.04
12.6	158.76
x=112.9	x2=1328.97

The sample mean is

Mean = (x / n) )

=8.3+7.3+13.8+12.6+11.1+9.5+13.7+14.2+9.8+12.6/10

=112.9/10

=11.29

Mean = 11.3

The sample standard is S

  S =( x2 ) - (( x)2 / n ) n -1

=1328.97-(112.9)210/9

=1328.97-1274.641/9

=54.3299

=6.0366

=2.4569

The sample standard = 2.4

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,9 =2.797  

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

= 2.797 * (2.4 / 10)

= 1.7

Margin of error = 1.7

The 95% confidence interval estimate of the population mean is,

- E < < + E

11.3 - 1.7 < < 11.3 +1.7

9.60 < < 13.00

Option 9.60 to 13.00 min is correct