Question

The owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on weekdays and weekends. The owner samples 24 weekdays and finds an average of 242.462 gallons of milk sold on those days with a standard deviation of 36.143. 28 Saturdays and Sundays are sampled and the average number of gallons sold is 335.717 with a standard deviation of 43.83. If a 95% confidence interval is calculated to estimate the difference between the average number of gallons sold on weekdays and weekends, what is the margin of error? Assume both population standard deviations are equal.

Question 7 options:

	1) 22.615
	2) 11.25930199
	3) 2.00855911
	4) 22.068
	5) 18.87

Answer 1

n1 = 24 , n2 = 28

s1 = 36.143 ,s2 = 43.83

Sp = sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))
= sqrt(((24-1) *36.143^2 + ( 28-1) * 43.83^2)/ (24+28 -2))
= 40.4757

t value at 95% = 2.009

Margin of error = t * Sp * sqrt(1/n1+1/n2)
= 2.009 * 40.4757*sqrt(1/24+1/28)
= 22.615

Option 1)