Mathematics, 31.07.2019 18:20 lolomgwtfnvm4

An article a florida newspaper reported on the topics that teenagers most want to discuss with their parents. the findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. these and other percentages were based on a national sampling of 531 teenagers. estimate the proportion of all teenagers who want more family discussions about school. use a 99% confidence level.

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From the information given, the 99% confidence interval estimate of the proportion of all teenagers who want more family discussions about school is (0.316, 0.424).

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The information given are as follows:

Sample size of n = 531

.Sample proportion of $\pi = 0.37$ .

99% confidence level

So $\alpha = 0.99$ , z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$ , so z = 2.575 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 - 2.575\sqrt{\frac{0.37(0.63)}{531}} = 0.316$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 + 2.575\sqrt{\frac{0.37(0.63)}{531}} = 0.424$

The 99% confidence interval estimate of the proportion of all teenagers who want more family discussions about school is (0.316, 0.424).

A similar problem is given at link

Answer from: dee3874

Confidence Interval for the proportion of all teenagers who want more family discussions about school is ( 31.6% , 42.4% )

Step-by-step explanation:

Given:

Number of teenagers in the sample , n = 531

Percentage of people like more discussion about family's financial situation = 46%

Percentage of people like more discussion about school, $\hat{p}$ = 37%

Percentage of people like more discussion about religion = 30%

To find: Confidence Interval for the proportion of all teenagers who want more family discussions about school

Level of confidence, α = 100 - 99 = 1% = 0.01

We know that

Critical Value for given level of significance, z_c = 2.58

The Standard Deviation,

$\sigma_{\hat{p}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.37\times(1-0.37)}{531}}=0.021$

The Required Confidence Interval is,

$\hat{p}\pm z_c.\sigma_{\hat{p}}=0.37\pm(2.58\times0.021)=0.37\pm0.0542$

=(0.37-0.0542,0.37+0.542)=(0.316,0.424)

Hence the Require confidence interval is ( 0.316 , 0.424 ).

Therefore, Confidence Interval for the proportion of all teenagers who want more family discussions about school is ( 31.6% , 42.4% )

Answer from: Quest

step-by-step explanation:

256/6=42

Answer from: Quest

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

$4/2 = 2\\7/3.5=2\\8/4=2\\10/5=2$

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

$y=\frac{x}{2}$

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.

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