Mathematics: A group of statistics students decided to conduct a survey at their university to estimate the average (mean) amount of time students spent studying per week. They sampled 554 students and found a mean of 22.3 hours per week. Assuming a population standard deviation of six hours, what is the 95% level of confidence? A. [16.3, 28.3] B. [21.64, 22.96] C. [21.80, 22.80] D. [20.22, 22.0]

A group of statistics students decided to conduct

AnaiyaKirksey8

22.01.2022

C. [21.80, 22.80]

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.025 = 0.975 , so z = 1.96

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{6}{\sqrt{554}} = 0.5$

The lower end of the interval is the sample mean subtracted by M. So it is 22.3 - 0.5 = 21.8 hours.

The upper end of the interval is the sample mean added to M. So it is 22.3 + 0.5 = 22.8 hours

So the correct answer is:

C. [21.80, 22.80]

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coolkiddKC

20.03.2023

A=28.26 in²

Step-by-step explanation:

We are given the formula for the area of a circle which is:

$A=\pi r^2$

Which means:

Area = π*radius of circle²

The radius of a circle is the length from the center of the circle to the edge of the circle.

The image tells us the radius of the circle is 3 in and the questions asks us to use 3.14 for π, therefore...

A=3.14*3²

A=3.14*9

A=28.26 in²

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