In: Statistics and Probability
A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. *
Solution :
Given that,
Point estimate = sample mean =
= 114.50
Population standard deviation =
= 12.30
Sample size = n =42
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 * ( 12.30/ 42
)
= 4.41
At 98% confidence interval
is,
- E <
<
+ E
114.50- 4.41 < < 114.50+ 4.41
(110.09 ,118.91)