Question

Michelle is brought in by a provider to determine the number of times per day their customers interact with their smartphone. You take a random sample of 20 customers and find the mean average number of interactions to be 45 times per day with a sample standard deviation of 8.

A. Construct a 95% confidence interval (to 2 decimal places) estimate for the population mean number of interactions.

B. What is the margin of error for the confidence interval you found in part b?

Answer 1

Solution :

Given that,

n = 20

$\bar x$ = 45

s = 8

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

Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.05 $\slash$ 2 = 0.025

Also, d.f = n - 1 = 20 - 1 = 19

$\therefore$  
ta/2.d.f-
  =  
ta/2.n-1
  =  _0.025,19 = 2.093

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

The margin of error is given by

E =  $\alpha$/2,d.f. * ( / $\sqrt{}$ n )

= 2.093 * ( 8/ $\sqrt{}$ 20 )

= 3.744

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

($\bar x$ - E ) <  $\mu$ <  ($\bar x$ + E)

( 45 - 3.744 )   <  $\mu$ <  ( 45 + 3.744 )

41.26 <  $\mu$ < 48.74

Required 95% confidence interval is ( 41.26 , 48.74 )

The margin of error for the confidence interval is 3.744