A random sample of 60 mathematics majors spent an average of $200.00 for textbooks for a term, with a standard deviation of $22.50. A random sample of 40 English majors spent an average of $180.00 for textbooks that term, with a standard deviation of $18.30. Calculate a 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors), assuming equal population variances.

20 +/- $6.74

= ( $13.26, $26.74)

The 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors) is ( $13.26, $26.74)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x1-x2 +/- margin of error

x1-x2 +/- z(√(r1^2/n1 + r2^2/n2)

Given that;

Mean x1 = $200

x2 = $180

Standard deviation r1 = $22.50

r2 = $18.30

Number of samples n1 = 60

n2 = 40

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

$200-$180 +/-1.645(√(22.5^2/60 +18.3^2/40)

$20 +/- 6.744449847374

$20 +/- $6.74

= ( $13.26, $26.74)

The 90% confidence interval for the difference in average amounts spent on textbooks (math majors - English majors) is ( $13.26, $26.74)

the answer is 31.27mph