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Assoli18 [71]
2 years ago
15

1. A researcher is interested in studying whether or not listening to music while jogging makes people run faster. He thinks tha

t listening to music will make people run faster. Luckily, he knows that, for the population he interested in (runners in Washington, DC), the mean running speed (μ) is 6mph, and the standard error is 2mph. He collects data from a sample of runners that only listen to music, and finds they have a mean running speed (M) of 9mph. No Sample Size givenA. State the hypothesisH0:H1:B. The researcher would like to conduct a One-Sample Z-test. Please calculate the Z-statistic (Z-obtained):
Mathematics
1 answer:
castortr0y [4]2 years ago
4 0

Step-by-step explanation:

1.Assuming the same sample size and considering the same value for the errors ( not taking into consideration the type of music, the volume of the sound and cow familiar the runner is with that type of stimuli, age group, time of the day/ number of days, running conditions like wether and equipment, distance) one can state:

A. Music has no influence over the running speed ( when jogging) in Washington DC

B.When listening to music, people ( in Washington DC) run faster while jogging

The mean running speed is a simple, ponderate or other type of mean ( that takes into consideration the variations of speed at the beginning and by the end of the race?

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Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
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Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

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A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx

A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}

A = 2.525 -2.474+9.798 + 2.525 - 2.474

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