A survey published in the American Journal of Sports Medicine reported the num- ber of meters (m) per week swum by two groups of swimmers—those who competed exclusively in breaststroke and those who competed in the individual medley (which includes breaststroke). The number of meters per week practicing the breaststroke 1 was recorded for each swimmer, and the summary statistics are given below. Is there sufficient evidence to indicate that the average number of meters per week spent prac- ticing breaststroke is greater for exclusive breaststrokers than it is for those swimming individual medley?

see explaination

Step-by-step explanation:

Let mu1 be the mean for exclusively breaststroke

Let mu2 be the mean for individual medley

(a) H o: mu1=mu2 (i.e. null hypothesis)

Ha: mu1> mu2 (i.e. alternative hypothesis)

(b) Since both sample sizes are grearter than n=30, we can use normal distriubtion.

It is a one-tailed test.

Given a=0.01, the critical value is Z(0.01) =2.33 (from standard normal table)

So the rejection region is Z>2.33

(c) The test statistic is

Z=(xbar2-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(9017-5853)/sqrt(7162^2/130+1961^2/80)

=4.76

(d) Since Z=4.76 is larger than 2.33, we reject the null hypothesis.

(e) In conclusion there is sufficient evidence to indicate that the average number of meter per week spent practicing breaststroke is greater for exclusive breaststrokers than it is for those swimming individual medley

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

~just a girl in love with shawn mendes