Answered

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Arnold wants to see if people who complete his weight-training program see an average change in their percent body fat. To test this claim, he measures the percent body fat of 30 random people who take his program. At the beginning of the program, the participants have an average of 32.8% body fat. At the end of the program, the average 27.7%. The differences in percent body fat is normally distributed with a standard deviation of 5.6%. Assuming the sample was random and observations were independent, run the appropriate test to analyze this data. Given: t*=2.045

Sagot :

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Hypotheses:

[tex]H_0: \mu_1 = \mu_2 \\\\H_a: \mu_1 \neq \mu_2\\[/tex]

Test statistic:

[tex]t= \frac{\bar{X}-\bar{Y}}{\frac{s_d}{\sqrt{n}}}[/tex]

  [tex]=\frac{32.8–27.7}{\frac{5.6}{\sqrt{30}}}\\\\=4.988[/tex]

Critical value: [tex]t* = 2.045[/tex]

[tex]t > 2.045[/tex]

Reject [tex]H_o[/tex] then

The data shows their weight training course the overall difference in their percentage the fats.

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