Question

When engaging in weight-control (fitness/fat burning) types of exercise, a person is expected to attain about 60% of their maximum heart rate. For 20-year-olds, this rate is approximately 120 bpm. A simple random sample of one hundred 20-year-olds was taken, and the sample mean was found to be 107 bpm with a standard deviation of 45 bpm. Researchers wonder if this is evidence to conclude that the expected level is actually lower than 120 bpm.
a) Test the following hypotheses at 5% significance level – follow the five steps.
b) Construct a 95% confidence interval for the population mean weight control heart rate for 20-year-olds and give an interpretation for the calculated confidence interval.
c) Do you think if we changed the confidence level to be 99%, the confidence interval would be wider or narrower? And why?

Answer 1

Answer:

a) We reject  H₀ we can conclude that the expected level is actually lower than 120 bpm

b) In the procedure

c) Confidence interval become wider

Step-by-step explanation:

1.- Hypothesis        Null hipothesis  H₀    μ₀  = 120

                      Alternate hipothesis  Hₐ    μₐ  < 120

2.- Is requiered significant level of 5 %   that means α = 0,05

and confidence interval is then   1  -  0,05  =  0.95    or 95 %

That definiton divide the area under the bell shape curve in two, the rejection area which is 0,05; and the acceptance region for hipothesis  H₀  95 % .

3.- According to significant level 5% the z value [ z(c) = -1.64 ]

4.- Now we calculate the vale for z(e) = z statistic

z (e) = ( x - μ₀ )/ (σ/√n)    ⇒ z (e) = (107-120)/ (45/√100)  ⇒ z(e) = - 13*10/45

z (e)  = - 2.88

5.- z (e) < z(c)  z(e) is in the rejection region

6.- We reject   H₀

7.- We have enough evidende to reject  H₀ we can conclude that the expected level is actually lower than 120 bpm

c) If we change the confidence level to 99 %   α = 0.01

now we have α = 0.01 and the confidence interval is 99 % the confidence interval become wider  z (c) in this case is  - 2.32 . Still z(e) < z(c)

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