Mathematics, 09.12.2019 20:31, hannahbear3747

Asurvey shows that college students spend on average about $20 per month on yum good ramen. a. assume that the survey was conducted on 200 students and assume a standard deviation of $6. find the 95% confidence interval of the estimate. b. assume that the survey was conducted on 200 students and assume a standard deviation of $3. find the 95% confidence interval of the estimate. c. assume that the survey was conducted on 400 students and assume a standard deviation of $6. find the 95% confidence interval of the estimate.

Answers: 3

Show answers

Answers

answered: samy14

a) The 95% confidence interval would be given by (19.168;20.832)

b) The 95% confidence interval would be given by (19.182;20.416)

c) The 95% confidence interval would be given by (19.412;20.588)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$20-1.96\frac{6}{\sqrt{200}}=19.168$

$20+1.96\frac{6}{\sqrt{200}}=20.832$

So on this case the 95% confidence interval would be given by (19.168;20.832)

Part b

For this case we just need to change the value for the deviation

$20-1.96\frac{3}{\sqrt{200}}=19.584$

$20+1.96\frac{3}{\sqrt{200}}=20.416$

So on this case the 95% confidence interval would be given by (19.182;20.416)

Part c

For this case we change again the deviation and the sample size.

$20-1.96\frac{6}{\sqrt{400}}=19.412$

$20+1.96\frac{6}{\sqrt{400}}=20.588$

So on this case the 95% confidence interval would be given by (19.412;20.588)

answered: Guest

step-by-step explanation:

73-22=51 the mass number minus the amount of protons gives you the amount of neutrons

answered: Guest

the answer for this question is : 360°

step-by-step explanation:

we have been given a triangle we need to tell which one of these is a triangle below to map onto itself.

360° rotation is the complete rotation which eventually leads to the same figure which is given.

the correct will be, that is 360° that will cause the triangle below to map onto itself.

Other questions on the subject: Mathematics

Mathematics, 21.06.2019 19:00, L3Zenith

What is not true about kl and mn? what is mn? (show )

Answers: 2

continue

Mathematics, 21.06.2019 20:30, shonesam98

Apinhole camera is made out of a light-tight box with a piece of film attached to one side and pinhole on the opposite side. the optimum diameter d(in millimeters) of the pinhole can be modeled by d=1.9[ (5.5 x 10^-4) l) ] ^1/2, where l is the length of the camera box (in millimeters). what is the optimum diameter for a pinhole camera if the camera box has a length of 10 centimeters

Answers: 1

continue

Mathematics, 21.06.2019 23:00, serenityburke

The areas of two similar triangles are 72dm2 and 50dm2. the sum of their perimeters is 226dm. what is the perimeter of each of these triangles?

Answers: 1

continue

Mathematics, 22.06.2019 00:00, daphnevlogs11

Which diagram shows lines that must be parallel lines cut by transversal?

Answers: 3

continue

Do you know the correct answer?

Asurvey shows that college students spend on average about $20 per month on yum good ramen. a. assum...