Question

My subject is Statistics. The homework that I am currently doing revolves around Confidence Intervals for Proportions. The last three questions require you to look at the first four questions to get your information. I forgot how to do this and I really need help completing it. Please help me! Thank you

Answer 1

Hello there. To solve this question, we'll have to remember some properties about probabilities and confidence intervals.

1. A survey of 1500 US teenagers found that 620 recycles regularly.

In this case, we find the proportion of how many teenagers from the survey recycles regularly by taking the ratio:

`\hat{p}=\frac{\text{teenagers that recycles}}{\text{sample of t}eenagers}`

In this case, we get:

`\hat{p}=(620)/(1500)`

Which evaluates to:

`\hat{p}\approx0.4133`

The proportion for the amount of teenagers that doesn't recycle is given by:

`\hat{q}=1-\hat{p}`

Which is approximately equal to:

`\hat{q}\approx1-0.4133\approx0.5867`

2. A recent study found that out of 4000 US teenagers, 2456 were overweight.

We find p, the proportion of how many teenagers are overweight taking the ratio:

`\hat{p}=\frac{overweight\text{ t}eenagers}{sample}`

Which is given by:

`\hat{p}=(2456)/(4000)`

This fraction is equal to:

`\hat{p}=0.614`

The amount of teenagers that are not overweight will then be given by:

`\hat{q}=1-\hat{p}\approx1-0.614=0.386`

3. A survey indicated that out of 3500 US high school students, 356 were identified as vegetarian or vegan.

We calculate the proportion of vegetarian/vegan students as:

`\hat{p}=(356)/(3500)\approx0.101`

And the amount of students who are not vegetarian/vegan as:

`\hat{q}=1-\hat{p}\approx1-0.101\approx0.899`

4. In a survey of 4300 US teachers, 2986 believe they will teach until they retire.

The proportion of teaches who believe they'll keep teaching until retirement is given by:

`\hat{p}=(2986)/(4300)\approx0.6944`

And the proportion of teacher who doesn't believe it is:

`\hat{q}=1-\hat{p}\approx1-0.6944\approx0.3056`

5. We need to build a 90% interval confidence using the data from question 1.

First, this interval is built as:

`\hat{p}\pm Z\sqrt{\frac{\hat{p}\cdot\hat{q}}{n}}`

Where n is the sample size, p and q are the found proportions and Z is the score we can find using a table.

This Z value will be found by searching for this value in a proper z-table:

`(0.9)/(2)=0.45`

That is, the percent of confidence divided by 2.

The corresponding Z value is 1.91, therefore we get the confidence interval as:

`0.4133\pm1.91\cdot\sqrt[]{(0.4133\cdot0.5867)/(1500)}`

Using a calculator, we get approximately:

`0.4133\pm0.0243`

Therefore the interval of confidence is:

`(0.389,0.4376)`