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2 years ago

1250+27.50x=1400+20x

Mathematics

1 answer:

never [62]2 years ago

5 0

Answer:

x = 20

Step-by-step explanation:

Simplifying

1250 + 27.50x = 1400 + 20x

Solving

1250 + 27.50x = 1400 + 20x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-20x' to each side of the equation.

1250 + 27.50x + -20x = 1400 + 20x + -20x

Combine like terms: 27.50x + -20x = 7.5x

1250 + 7.5x = 1400 + 20x + -20x

Combine like terms: 20x + -20x = 0

1250 + 7.5x = 1400 + 0

1250 + 7.5x = 1400

Add '-1250' to each side of the equation.

1250 + -1250 + 7.5x = 1400 + -1250

Combine like terms: 1250 + -1250 = 0

0 + 7.5x = 1400 + -1250

7.5x = 1400 + -1250

Combine like terms: 1400 + -1250 = 150

7.5x = 150

Divide each side by '7.5'.

x = 20

Simplifying

x = 20

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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

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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 = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

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

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

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n = $54.79^{2}$

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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%.

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Answer:

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Step-by-step explanation:

We are given the following situation in the question:

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This is a n example of statistical Inference.

Statistical inference is the procedure of making inference or estimating parameters of a population with the help of test statistics.

In the given situation the university students are the population and the students of a particular class are sample.

The professor with the help of sample statistic (mean age of students in class) approximated the mean age of students in the university.

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