A person read that the average number of hours an adult sleeps on Friday night to Saturday morning was 7.2 hours. The

researcher feels that college students do not sleep 7.2 hours on average. The researcher randomly selected 15 students and

found that on average they slept 8.3 hours. The standard deviation of the sample is 1.2 hours. If you wanted to test this claim,

what type of statistical test would you do and why?

Calculate the value of z and its probability, since with this we can know how likely it is that this will happen.

We have that the mean (m) is equal to 8.3, the standard deviation (sd) 1.2 and the sample size (n) = 15

They ask us for P (x =7.2)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (7.2 - 8.3) / (1.2 / (15 ^ (1/2))

z = -3.55

With the normal distribution table (attached), we have that at that value the approximate probability is:

P (z = -3.55) = 0.0001

The probability is 0.01 %

This affirms that the students do not sleep what the study says because the probability of this happening according to the survey is almost nil.

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step-by-step explanation:

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$3720

step-by-step explanation:

taxable income = $31000

as per the 2018-2019 us tax bracket, the tax rate is 12% for an individual whose taxable income is between $9,526 to $38,700.

so, the tax to be paid by him = 12% of 31000

= 0.12 * 31000     [12% converted to decimal as 0.12]

= $3720