The AARP (American Association of Retired People) report that at least 60% of retired people under the age of 65 would return to work on a full-time basis if a suitable job were available. A sample of 500 retirees under the age of 65 showed that 315 would return to work. Can we conclude that more than 60% would return to work? Test at the 2% level of significance.

Step-by-step explanation:

Proportion of retired people under the age of 65 would return to work on a full-time basis if a suitable job were available = 60/100 = 0.6 = P

Null hypothesis: P ≤ 0.6

Alternative: P > 0.6

First, to calculate the hypothesis test, lets workout the standard deviation

SD =  √[ P x ( 1 - P ) / n ]

where P = 0.6, 1 - P = 0.4, n = 500

SD = √[ (0.6 x 0.4) / 500]

SD = √ (0.24 / 500)

SD = √0.00048

SD = 0.022

To calculate for the test statistic, we have:

z = (p - P) / σ where p = 315/500 = 0.63, P = 0.6, σ = 0.022

z = (0.63 - 0.6) / 0.022

z = 0.03/0.022

z = 1.36

At the 2% level of significance, the p value is less than 98% confidence level, thus we reject the null hypothesis  and conclude that more than 60% would return to work.

the answer is d. (1.2, -4.7) on edgn

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geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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