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I attempted to solve the following statistics questions below and I was wondering if someone would be willing to check my answers and confirm they're correct? I was also wondering if there is a way I could expand my answer for c).

According to an article in the New York Times, America's elderly paid an average of 3757 dollars for healthcare in 2003. The figure was based on a random sample of 560 elderly Americans and the standard deviation of the sample was 870 dollars.

(a) Determine an 88% confidence interval for the actual average spending on health care by elderly Americans in 2003.

z=(1-0.88)/2 = 0.06

1-0.06 = 0.94

p(z < 1.56) = 0.94

x ± z(s/√n)

3757 ± ((1.56)(870/√560))

3757 - 57.35216 = 3699.65

3757 + 57.35216 = 3814.35

Therefore the confidence interval is between 3699.65 and 3814.35.

(b) Determine a 94% confidence interval for the actual average spending on health care by elderly Americans in 2003.

z=(1-0.94)/2 = 0.03

1-0.03 = 0.97

p(z < 1.89) = 0.97

x ± z(s/√n)

3757 ± ((1.89)(870/√560))

3757 - 69.48435 = 3687.52

3757 + 69.48435 = 3826.48

Therefore the confidence interval is between 3687.52 and 3826.48

(c) Interpret the confidence interval constructed in part (b)

94% of elder americans will pay between 3687.52 and 3826.48 in health care in 2003.

I feel like I'm missing something, but I don't really know how to interpret the data further than that?

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Part (a) and part (b) are correct. But I think in part (c) the 94% should refer to the confidence of the money falling in the confidence interval

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  • $\begingroup$ Welcome to Math.SE! You might want to consider elaborating on what a correct interpretation of the confidence interval would be so that your answer might be more helpful to future readers. $\endgroup$ – Brian Apr 2 at 1:50

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