Question

A survey of several 10 to 12 year olds recorded the following amounts spent on a trip to the mall: $14.73,$18.13,$11.20,$14.89,$18.78,$21.08,$17.26 Construct the 99% confidence interval for the average amount spent by 10 to 12 year olds on a trip to the mall. Assume the population is approximately normal.

Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to two decimal places.

Answer 1

Solution:

x	x2
14.73	216.9729
18.13	328.6969
11.2	125.44
14.89	221.7121
18.78	352.6884
21.08	444.3664
17.26	297.9076
x=116.07	x2=1987.7843

The sample mean is

Mean = (x / n) )

=14.73+18.13+11.2+14.89+18.78+21.08+17.26/7

=116.07/7

=16.5814

The sample standard is S

  S =( x2 ) - (( x)2 / n ) n -1

=1987.7843-(116.07)27/6

=1987.7843-1924.6064/6

=63.1779/6

=10.529/6

=3.2449

Degrees of freedom = df = n - 1 = 7 - 1 =6

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,6 =3.707

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

=3.707 * (3.24 / 7)

= 4.54

Margin of error = 4.54

The 99% confidence interval estimate of the population mean is,

- E < < + E

16.58 -4.54< < 16.58 + 4.54

12.04 < < 21.12

(12.04, 21.12 )