A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation...

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. *

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Expert Solution

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 * ( / $\sqrt$ n)

= 2.326 * ( 12.30/ $\sqrt$ 42 )

= 4.41
At 98% confidence interval
is,

- E < < + E

114.50- 4.41 < < 114.50+ 4.41

(110.09 ,118.91)

orchestra answered 1 year ago

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