Answer:
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)
Answer:
6.48 dollars total,
Step-by-step explanation:
First, you would want to multiply the amount of money per pound, by the amount of pounds. 1.60 multiplied by 4 1/2 or 4.5, would equal to 7.2 dollars. In order to find 10% of 7.2, your would need to multiply 7.2 by 0.10, which would give you 0.72. Then you would subtract 7.2 by 0.72 to get 6.48. So your answer would be 6.48 dollars. Hoped this helped.
The answer to m2=91 is 40.5
Answer:
Response variable is another term for dependent variable. So the response variable is y.
