Question

Justin is interested in buying a digital phone. He visited 11 stores at random and recorded the price of the particular phone he wants. The sample of prices had a mean of 188.08 and a standard deviation of 10.94. (a) What t-score should be used for a 95% confidence interval for the mean, μ, of the distribution? t* = equation editorEquation Editor (b) Calculate a 95% confidence interval for the mean price of this model of digital phone: (Enter the smaller value in the left answer box.)r

Answer 1

Solution

Note that, Population standard deviation() is unknown. So we use t distribution.

a)

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 11 - 1 = 10

    =    =  0.025,10 = 2.228

i.e. t* = 2.228

( use t table or t calculator to find this value..)

b)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.228 * (10.94 / 11)

= 7.3491

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(188.08 - 7.3481 )   <   <  (188.08 + 7.3491)

180.73 <   < 195.43

Required 95% confidence interval is

(180.73 , 195.43)