Question

In: Statistics and Probability

A fitness trainer claims that high intensity power training decreases the body fat percentages of females....

A fitness trainer claims that high intensity power training decreases the body fat percentages of females. The table below shows the body fat percentages of

eight

females before and after ten weeks of high intensity power training. At

alphaαequals=0.10

is there enough evidence to support the​ trainer's claim? Assume the samples are random and​ dependent, and the population is normally distributed. Complete parts​ (a) through​ (e) below.

​(a) Identify the claim and state

Upper H0

and

Upper H Subscript aHa.

What is the​ claim?

High intensity power training

decreases

the body fat percentages of females.

Let

mu Subscript dμd

be the hypothesized mean of the difference in the body fat percentages

​(beforeminus−​after).

What are

Upper H0

and

Upper H Subscript aHa​?

​(b) Find the critical​ value(s) and identify the rejection​ region(s).

Select the correct choice below and fill in any answer boxes to complete your choice.

​(Round to three decimal places as​ needed.)

​(c) Calculate

d overbard

and

s Subscript dsd.

d overbardequals=_____

​(Round to three decimal places as​ needed.)

Calculate

s Subscript dsd.

s Subscript dsdequals=_____

​(Round to three decimal places as​ needed.)

​(d) Find the standardized test statistic t.

tequals=_____

​(Round to three decimal places as​ needed.)

​(e) Decide whether to reject or fail to reject the null hypothesis and interpret the decision in the context of the original claim.

(1)

the null hypothesis. There

(2)

enough evidence to

(3)

the claim that high intensity power training

(4)

the body fat percentages of females.

---

Female Body_Fat_%_(before) Body_Fat_%_(after)
1 27.2 27.1
2 23.6 23
3 23.9 23.8
4 23.5 23.4
5 24.2 24
6 27.4 26.8
7 28.3 28.2
8 28.1 28.5

Solutions

Expert Solution

From the sample data, it is found that the corresponding sample means are:

\bar X_1 = 25.775 ; \bar X_2 = 25.6

Also, the provided sample standard deviations are:

s1​=2.15 s2=2.276 and the sample size is n = 8. For the score differences we have \bar D = 0.175 ; sD = 0.32.

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD = 0

Ha: μD > 0

This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

Rejection Region

Based on the information provided, the significance level is α=.1, and the degrees of freedom are df = 7.

Hence, it is found that the critical value for this right-tailed test is tc= 1.415.

The rejection region for this right-tailed test is R={t : t > 1.415}.

Test Statistics

The t-statistic is computed as shown in the following formula:

Decision about the null hypothesis

Since it is observed that t = 1.549 > tc=1.415, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0827, and since p = 0.0827 < .1, it is concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the high intensity power training decreases the body fat % of females, at the .1 significance level.


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