Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 44 who smoke. Suppose a sample of 2653 Americans over 44 is drawn. Of these people, 2149 don't smoke.

A. Using the data, estimate the proportion of Americans over 44 who smoke. Enter your answer as a fraction or a decimal number rounded to three decimal places.

B. Using the data, construct the 99% confidence interval for the population proportion of Americans over 44 who smoke. Round your answers to three decimal places.

Lower endpoint: ______ Upper endpoint: ______

Consider the value of t such that 0.05 of the area under the curve is to the right of t.

Assuming the degrees of freedom equals 24, what is the t value

Answer #1

1)

A)

Sample proportion
= 2149 / 2653 = **0.810**

B)

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.810 - 2.576 * sqrt [ 0.810 * ( 1 - 0.810) / 2653] < p < 0.810 + 2.576 * sqrt [ 0.810 * ( 1 - 0.810) / 2653]

0.790 < p < 0.830

**Lower endpoint = 0.790**

**Upper endpoint = 0.830**

2)

We have to calculate t such that

P(T > t )= 0.05

With 24 df ,

**t = 1.711**

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