What is (0.5x + 15) + (8.2x- 16.6)

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

50 points decreased by 26% is how many points?

vampirchik [111]

The first one is 13 points and the 2nd one is 8. I'm pretty sure, hope this helps! :)

6 0

2 years ago

The Hamden board of education wants to know how the community feels about building a new media center for the middle school. The

Aleks [24]

According to the statement above, The Hamden board of education called every tenth person on the registration list. So let's analyze each case:

The sample is not randomly chosen (FALSE)

Given that the statement doesn't tell us anything about the way they choose the sample, it is reasonable to conclude that this is a randomly chosen. They called every tenth person on the registration list until the number of people was 40.

The sample should be larger to give more reliable information (TRUE)

You did not have to use mathematics to determine that you would need more information to get a conclusion. You must increase the sample, that is, the sample must be larger to give a reliable information.

The sample size is too large to make inferences (False)

This is explained in the previous item. If the sample should be lager is because the size is not too large.

The sample size is too small to represent the population (TRUE

This is true because 40 voters represent barely 0.5% of the entire list. This list has 7300 voters, so getting the conclusion from this sample doesn't provide with a strong conclusion.

The sample size is too small and will show larger variation. (FALSE)

Although the sample size is too small, the sample size not necessarily will show variation. In fact, it is possible that it does not show any variation and most of the people feel well about building a new media center for the middle school but it doesn't mean that the whole community does.

The sample is invalid because it randomly chooses voters. (FALSE)

It is false because in probability studies the sample is chosen randomly, so you get conclusions about the whole population always taking samples that represent the population as a whole.

The sample size is too small and can lead to false inferences (TRUE)

You can get false conclusions given that the sample size is too small. It's important to note that the sample size supports the conclusion of the study, so the sample must increase to have a reliable study.

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Find what is common between 35 and 55....and that would be 5....so factor 5 out

35x + 55 =
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