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The Consumer Reports National Research Center conducted a telephone survey of 2000 adults to learn about the major economic concerns for the future (Consumer Reports, January 2009). The survey results showed that 1760 of the respondents think the future health of Social Security is a major economic concern.

a. What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern (to 2 decimals)?
b. At 90% confidence, what is the margin of error (to 4 decimals)?
c. Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern (to 3 decimals).(, )
d. Develop a 95% confidence interval for this population proportion (to 4 decimals).

Respuesta :

Answer:

a) [tex] \hat p =\frac{X}{n}=\frac{1760}{2000}=0.88[/tex]

b) [tex]ME=1.64* \sqrt{\frac{0.88*(1-0.88)}{2000}}=0.0119[/tex]

c) [tex]0.88 - 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.868[/tex]

[tex]0.88 + 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.892[/tex]

And the 90% confidence interval would be given (0.868;0.892).

d) [tex]0.88 - 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8658[/tex]

[tex]0.88 + 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8942[/tex]

And the 90% confidence interval would be given (0.8658;0.8942).

Step-by-step explanation:

Part a

For this case the point of estimate for the population proportion is given by:

[tex] \hat p =\frac{X}{n}=\frac{1760}{2000}=0.88[/tex]

Part b

The confidence interval would be given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.64[/tex]

The margin of error is given by:

[tex] ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

And if we replace we got:

[tex]ME=1.64* \sqrt{\frac{0.88*(1-0.88)}{2000}}=0.0119[/tex]

Part c

And replacing into the confidence interval formula we got:

[tex]0.88 - 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.868[/tex]

[tex]0.88 + 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.892[/tex]

And the 90% confidence interval would be given (0.868;0.892).

Part d

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.96[/tex]

[tex]0.88 - 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8658[/tex]

[tex]0.88 + 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8942[/tex]

And the 90% confidence interval would be given (0.8658;0.8942).

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