Question

Suppose Laura, a facilities manager at a health and wellness company, wants to estimate the difference in the average amount of time that men and women spend at the company's fitness centers each week.

Laura randomly selects 14 adult male fitness center members from the membership database and then selects 14 adult female members from the database. Laura gathers data from the past month containing logged time at the fitness center for these members. She plans to use the data to estimate the difference in the time men and women spend per week at the fitness center. The sample statistics are summarized in the table.

Population	Population description	Population mean (unknown)	Sample size	Sample mean (min)	Sample standard deviation (min)
1	male	?1	?1=14	?⎯⎯⎯1=120.1	?1=40.3
2	female	?2	?2=14	?⎯⎯⎯2=104.5	?2=25.9

df=22.174

The population standard deviations are unknown and unlikely to be equal, based on the sample data. Laura plans to use the two-sample ?-procedures to estimate the difference of the two population means, ?1−?2.

Compute the margin of error, ?, for the 95% confidence interval for the difference of the population means using software or a table of ?-distributions. If you are using software, you may find some software manuals helpful. Provide your answer precise to at least one decimal place.

m=

Answer 1

margin of error=26.5 and confidence interval=(-42.10,-10.90)

difference of two sample mean=1-2=120.1-104.5=15.6

standard error of difference of two sample means=SE(difference)=sqrt(s1²/n1+s2²/n2)=

=sqrt(40.3*40.3/14+25.9*25.9/14)=12.80

margin of error=t(0.05/2,df)*SE(difference) =t(0.05/2,22.174)*12.80=2.07*12.80=26.50

(1-alpha)*100% confidence interval for population mean difference=

=sample mean difference±t(alpha/2,df)*SE(difference)=sample mean difference±margin of error

95% confidence interval =15.6±26.50=(-42.10,-10.90)