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:
[tex]\hat{p}=\frac{\text{teenagers that recycles}}{\text{sample of t}eenagers}[/tex]In this case, we get:
[tex]\hat{p}=\frac{620}{1500}[/tex]Which evaluates to:
[tex]\hat{p}\approx0.4133[/tex]The proportion for the amount of teenagers that doesn't recycle is given by:
[tex]\hat{q}=1-\hat{p}[/tex]Which is approximately equal to:
[tex]\hat{q}\approx1-0.4133\approx0.5867[/tex]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:
[tex]\hat{p}=\dfrac{overweight\text{ t}eenagers}{sample}[/tex]Which is given by:
[tex]\hat{p}=\frac{2456}{4000}[/tex]This fraction is equal to:
[tex]\hat{p}=0.614[/tex]The amount of teenagers that are not overweight will then be given by:
[tex]\hat{q}=1-\hat{p}\approx1-0.614=0.386[/tex]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:
[tex]\hat{p}=\dfrac{356}{3500}\approx0.101[/tex]And the amount of students who are not vegetarian/vegan as:
[tex]\hat{q}=1-\hat{p}\approx1-0.101\approx0.899[/tex]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:
[tex]\hat{p}=\dfrac{2986}{4300}\approx0.6944[/tex]And the proportion of teacher who doesn't believe it is:
[tex]\hat{q}=1-\hat{p}\approx1-0.6944\approx0.3056[/tex]5. We need to build a 90% interval confidence using the data from question 1.
First, this interval is built as:
[tex]\hat{p}\pm Z\sqrt{\dfrac{\hat{p}\cdot\hat{q}}{n}}[/tex]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:
[tex]\frac{0.9}{2}=0.45[/tex]That is, the percent of confidence divided by 2.
The corresponding Z value is 1.91, therefore we get the confidence interval as:
[tex]0.4133\pm1.91\cdot\sqrt[]{\frac{0.4133\cdot0.5867}{1500}}[/tex]Using a calculator, we get approximately:
[tex]0.4133\pm0.0243[/tex]Therefore the interval of confidence is:
[tex](0.389,0.4376)[/tex]