Question

A software developer wants to know how many new computer games people buy each year. A sample of 2395 people was taken to study their purchasing habits. Construct the 98% confidence interval for the mean number of computer games purchased each year if the sample mean was found to be 6.3. Assume that the population standard deviation is 2.2. Round your answers to one decimal place.

Answer #1

Solution :

Given that,

Point estimate = sample mean =
= 6.3

Population standard deviation =
= 2.2

Sample size = n =2395

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z_{/2}
= Z_{0.01} = 2.326 ( Using z table ( see the 0.01 value in
standard normal (z) table corresponding z value is 2.326 )

Margin of error = E = Z/2
* (
/n)

= 2.326 * ( 2.2 / 2395
)

E= 0.1

At 98% confidence interval estimate of the population mean

is,

- E <
<
+ E

6.3 - 0.1 <
< 6.3+ 0.1

6.2 <
< 6.4

(6.2 , 6.4 )

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