Question

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?

Answer 1

Answer:

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

Answer 2

There is sufficient evidence to indicate that the mean number of breaststroke is greater for exclusive breaststrokers than for swimming individual medley.

What are null hypotheses and alternative hypotheses?

In null hypotheses, there is no relationship between the two phenomena under the assumption that it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Let μ₁ be the mean for the exclusive breaststroke.

Let μ₂ be the mean for the individual medley.

For the null hypothesis, we have

H₀: μ₁ = μ₂

For the alternative hypothesis, we have

Hₐ: μ₁ > μ₂

The sample size of both is greater than 30, then the normal distribution will be

a = 0.01

The critical value is z(0.01) = 2.33 (From standard normal table)

So the rejection region is z > 2.33

The test statistic will be

$z = \dfrac{\mu_2 - \mu_1}{\sqrt{s^2_1/n_1 + s^2_2/n_2}}\\\\\\z = \dfrac{9017 - 5853}{\sqrt{7162^2/130+1961^2/80}}\\\\\\z = 4.76$

Since the value of z is greater than 2.33, then we reject the null hypothesis.

More about the null hypotheses and alternative hypotheses link is given below.

https://brainly.com/question/9504281