How to calculate the rejection region (energy-drink consumption)

DownNAbbey

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The exercise provides answers but I am not reaching the same solution for the rejection region.



A researcher at the University of Washington medical school believes that energy-drink consumption may increase heart rate. Suppose it is known that heart rate is normally distributed with an average of 70 bpm (beats per minute) for adults. A random sample of twenty-five adults was selected and it was found that their average heartbeat after energy-drink consumption was 73 bpm, with a standard deviation of 7 bpm.

Test at the 10% significance level if we can infer that energy-drink consumption increases heart rate.




Answer:

\(\displaystyle H_0\, :\, \mu\, =\, 70,\, H_1\, :\, \mu\, >\, 70\)

Test statistic:

\(\displaystyle t\, =\, \dfrac{\bar{X}\, -\, \mu}{\left(\frac{s}{\sqrt{n\,}}\right)}\)

Rejection region:


\(\displaystyle t\, >\, t_{0.10,24}\, =\, 1.318\)

\(\displaystyle t\, =\, 2.143\)

Conclusion: Reject H0. Yes, we can infer that energy-drink consumption increases heart rate at the 10% level, according to this data.



I don't quite grasp the formula for finding the rejection region. Thanks for the help if you can.
 

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The exercise provides answers but I am not reaching the same solution for the rejection region.



A researcher at the University of Washington medical school believes that energy-drink consumption may increase heart rate. Suppose it is known that heart rate is normally distributed with an average of 70 bpm (beats per minute) for adults. A random sample of twenty-five adults was selected and it was found that their average heartbeat after energy-drink consumption was 73 bpm, with a standard deviation of 7 bpm.

Test at the 10% significance level if we can infer that energy-drink consumption increases heart rate.




Answer:

\(\displaystyle H_0\, :\, \mu\, =\, 70,\, H_1\, :\, \mu\, >\, 70\)

Test statistic:

\(\displaystyle t\, =\, \dfrac{\bar{X}\, -\, \mu}{\left(\frac{s}{\sqrt{n\,}}\right)}\)

Rejection region:


\(\displaystyle t\, >\, t_{0.10,24}\, =\, 1.318\)

\(\displaystyle t\, =\, 2.143\)

Conclusion: Reject H0. Yes, we can infer that energy-drink consumption increases heart rate at the 10% level, according to this data.



I don't quite grasp the formula for finding the rejection region. Thanks for the help if you can.

Please share your work with us so that we could assist you to grasp the concept.
 
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View attachment 8633

Please share your work with us so that we could assist you to grasp the concept.

No problem.

1 - .10 = .9

Then finding the value on the by reverse looking up the cumulative standardized normal probabilities table. Since it is between two values it is averaged.

1.28 + 1.29 = 2.57 / 2 = 1.285

So the rejection region is 1.285 but obviously I'm incorrect.
 
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