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

< 6 cm to > 26 cm i believe

step-by-step explanation:

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine