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.

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

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

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 [tex]n = 1664, \pi = \frac{939}{1664} = 0.5643[/tex]

95% confidence level

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

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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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