Mathematics, 19.03.2021 15:40, bry7474

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

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Answers

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.

answered: Guest

step-by-step explanation:

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answered: Guest

step-by-step explanation: