Question

The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He randomly samples 25 men and 28 women that play on his course. He finds the average age of the men to be 36.105 with a standard deviation of 5.507. The average age of the women was 48.973 with a standard deviation of 5.62. He uses this information to calculate a 90% confidence interval for the difference in means, (-15.434, -10.302). The best interpretation of this interval is which of the following statements?

Question 10 options:

1)	We are 90% sure that the average age difference between all males and females is between -15.434 and -10.302.

2)	We are 90% confident that the difference between the average age of all men and all women who play golf at the course is between -15.434 and -10.302

3)	We are certain that the difference between the average age of all men and all women is between -15.434 and -10.302.

4)	We do not know the population means so we do not have enough information to make an interpretation.

5)	We are 90% confident that the difference between the average age of the men and women surveyed is between -15.434 and -10.302

Disability Services introduced a new mentorship program to help students with disabilities achieve better scholastic results. Test grades were recorded for 21 students before and after the program was introduced. The average difference in test score (after - before) was -3.739 with a standard deviation of 2.454. If Disability Services is interested in creating a 95% confidence interval for the true average difference in test scores, what is the margin of error?

Question 11 options:

1)

1.0496

2)

1.117

3)

0.5355

4)

1.1136

5)

0.9215

A restaurant wants to test a new in-store marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 6 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The average difference in the sales quantity (after - before) is -63.688 with a standard deviation of 41.8989. Calculate a 90% confidence interval to estimate the true average difference in nationwide sales quantity before the ad campaign and after.

Question 12 options:

1)	(98.1557, -29.2203)

2)	(-80.7932, -46.5828)

3)	(-98.1557, -29.2203)

4)	(-96.9264, -30.4496)

5)	(-65.703, -61.673)

Answer 1

10)

We are 90% confident that the difference between the average age of all men and all women who play golf at the course is between -15.434 and -10.302

11)

sample std deviation s=	2.454
std error 'sx=s/√n=	0.5355

for 95% CI; and 20 df, value of t=		2.086
margin of error E=t*std error    =		1.117

12)

sample mean 'x̄=	-63.688
sample size   n=	6.00
sample std deviation s=	41.899
std error 'sx=s/√n=	17.1052

for 90% CI; and 5 df, value of t=		2.015
margin of error E=t*std error    =		34.467
lower bound=sample mean-E =		-98.1557
Upper bound=sample mean+E =		-29.2203
from above 90% confidence interval for population mean =(-98.1557 , -29.2203)