Question

The average expenditure on Valentine's Day was expected to be 100.89 (USA Today, February 13, 2006). Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 60 male consumers was 130.6, and the average expenditure in a sample survey of 35 female consumers was 63.24 . Based on past surveys, the standard deviation for male consumers is assumed to be 30 , and the standard deviation for female consumers is assumed to be 14 . The z   value is 2.576.
a. At 99% confidence, what is the margin of error?

b. Develop a 99% confidence interval for the difference between the two population means.

Answer 1

= Mean (male expenditure ) =130.6 , ₁ = 30 and n1 =60

=mean (female expenditure ) = 63.24 , ₂ =14 and n2 =35

A)

Margin of error = Z value* SE( - )

   =

Margin of error =  2.578 * 4.5387 = 11.70077

B)

99% confidence interval for the difference between the two population means.

( ( 130.6-63.24 )- margin of error ,  ( 130.6-63.24 ) +margin of error )

( 67.36 -  11.70077 ,  67.36 +  11.70077)

(  55.65923 ,  79.06077 )

99% confidence interval for the difference between the two population means is  (  55.65923 ,  79.06077 )

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Thank You !