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BigorU [14]
2 years ago
8

A report on consumer financial literacy summarized data from a representative sample of 1,664 adult Americans. Based on data fro

m this sample, it was reported that over half of U.S. adults would give themselves a grade of A or B on their knowledge of personal finance. This statement was based on observing that 939 people in the sample would have given themselves a grade of A or B.
(a) Construct and interpret a 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance. (Round your answers to three decimal places.)
(,)
(b) Is the confidence interval from part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of A or B? Explain why or why not.
Because this confidence interval _____ is entirely below 0.5contains 0.5is entirely above 0.5 , the interval ______ is not consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.
Mathematics
1 answer:
daser333 [38]2 years ago
8 0

Answer:

a. The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

b. Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Sample of 1,664 adult Americans, 939 people in the sample would have given themselves a grade of A or B in personal finance.

This means that n = 1664, \pi = \frac{939}{1664} = 0.5643

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 - 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.54

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5643 + 1.96\sqrt{\frac{0.5643*0.4357}{1644}} = 0.588

The 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is (0.54, 0.588). This means that we are 95% sure that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between these two bounds.

(b) Is the confidence interval from part (a) consistent with the statement that a majority of adult Americans would give themselves a grade of A or B?

Yes, because the confidence interval is entirely above 0.5.

Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

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ANSWER

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{t  + 6}{ t + 2}
where,
t \ne - 2



EXPLANATION

We want to simplify the rational expression

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }


We can observe that the numerator of the given rational expression is a quadratic trinomial and the denominator is a difference of two squares



We need to split the middle term in the numerator and rewrite the denominator as a difference of two squares to obtain,




\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{{t}^{2}  + 6t - 2t - 12}{ {t}^{2}  -  {2}^{2} }



We now factor to obtain,

\frac{ {t}^{2}  + 4t - 12}{ {t}^{2} - 4 }  =  \frac{t (t+ 6) - 2(t  + 6)}{ (t  -  2)(t + 2)}


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We now cancel out common factors to obtain,

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The restriction is that, the denominator cannot be zero.

Thus

t + 2 \ne0


This implies that,

t  \ne - 2
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