Question

In a survey of 2046 adults in a recent​ year, 742 made a New​ Year's resolution to eat healthier. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

The​ 90% confidence interval for the population proportion p is

Answer 1

Answer:

Given that,

n = 2046

x = 742

= x / n = 742 / 2046 = 0.362

1 - = 1 - 0.362 = 0.638

(a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * (((0.362 * 0.638) / 2046)

= 0.017

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.362 - 0.017 < p < 0.362 + 0.017

0.345 < p < 0.379

(0.345 , 0.379)

(b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.362 * 0.638) / 2046)

= 0.021

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.362 - 0.021 < p < 0.362 + 0.021

0.341 < p < 0.383

(0.341 , 0.383)

95% confidence interval is wider than 90% .

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