Question

A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Create a 99% confidence interval for the population proportion of passing test scores.

Enter the lower and upper bounds for the interval in the following boxes, respectively.

You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.

________

< p <

________

An online retailer wants to estimate the number of visitors that click on their advertisement from a particular website. Of 978 page views in a day, 8% of the users clicked on the advertisement.

Create a 90% confidence interval for the population proportion of visitors that click on the advertisement.

Enter the lower and upper bounds for the interval in the following boxes, respectively.

You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.

________

< p <

________

Answer 1

a)
`0.77 - 2.58\sqrt{(0.77(1-0.77))/(40)}=0.5983`

`0.77 + 2.58\sqrt{(0.77(1-0.77))/(40)}=0.9417`

The 99% confidence interval would be given by (0.5983;0.9417) and in a percentage would be:

59.83% < p< 94.17%

b)
`0.08 - 1.64\sqrt{(0.08(1-0.08))/(978)}=0.0658`

`0.08 + 1.64\sqrt{(0.08(1-0.08))/(978)}=0.0942`

The 90% confidence interval would be given by (0.0658;0.0942) and in a percentage would be:

6.58% < p< 9.42%

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".

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by
`\alpha=1-0.99=0.01` and
`\alpha/2 =0.005`. And the critical value would be given by:

`z_(\alpha/2)=-2.58, z_(1-\alpha/2)=2.58`

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

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.77 - 2.58\sqrt{(0.77(1-0.77))/(40)}=0.5983`

`0.77 + 2.58\sqrt{(0.77(1-0.77))/(40)}=0.9417`

The 99% confidence interval would be given by (0.5983;0.9417) and in a percentage would be:

59.83% < p< 94.17%

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`z_(\alpha/2)=-1.64, z_(1-\alpha/2)=1.64`

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

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.08 - 1.64\sqrt{(0.08(1-0.08))/(978)}=0.0658`

`0.08 + 1.64\sqrt{(0.08(1-0.08))/(978)}=0.0942`

The 90% confidence interval would be given by (0.0658;0.0942) and ina percentage would be:

6.58% < p< 9.42%