Question

A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 96 students who averaged 41.6 texts per day. The standard deviation was 13.3 texts.

Answer 1

Solution :

Given that,

Point estimate = sample mean = =41.6

Population standard deviation =    = 13.3

Sample size = n =96

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z/2    * ( /n)

= 2.326 * ( 13.3/  96 )

=3.1574
At 98% confidence interval estimate of the population mean
is,

- E < < + E

41.6 -3.1574   <   < 41.6 + 3.1574

38.4426 <   < 44.7574

( 38.4426 ,44.7574)