Question

In a survey, 29 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $48.1 and standard deviation of $2.6. Estimate how much a typical parent would spend on their child's birthday gift (use a 90% confidence level). Give your answers to 3 decimal places.

Express your answer in the format of ¯xx¯ ±±  E.
$ ±± $

Answer 1

Solution:

Given that,

n = 29

= 48.1   

s = 2.6  

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

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 29-1 = 28

    =    =  _0.05,28 = 1.7011

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.7011 * ( 2.6 / 29 )

= 0.8213

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

±  E

48.1 ± 0.8213

Answer : $  48.1 ± $ 0.821