Question

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]

Answer 1

Answer:

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]

Answer 2

Answer:

C. [21.80, 22.80]

Step-by-step explanation:

We have a sample of size n=554 taken from a population.

The mean os fthis sample is 22.3.

We know that the population's standard deviation is 6.

We need to calculate a 95% confidence interval.

For a 95% confidence interval, the z-value is z=1.96.

The margin of error can be calculated as:

$E=z\sigma/\sqrt{n}=1.96*6/\sqrt{554}=11.76/23.54=0.5$

This can be calculated as:

$M-z\sigma/\sqrt{n}\leq\mu\leq M+z\sigma/\sqrt{n}\\\\22.3-0.5\leq\mu\leq 22.3-0.5\\\\21.8\leq\mu\leq 22.8$

The 95% CI is

$21.8\leq\mu\leq 22.8$