Question

In: Math

In a March 26-28, 2010 Gallup poll, 473 respondents (out of 1033) approved of they way...

In a March 26-28, 2010 Gallup poll, 473 respondents (out of 1033) approved of they way President Obama handled the health care debate (regardless of their opinions on health care policy).

a) What is your point estimate for this proportion?

b) Construct a 95% confidence interval for this opinion. did a significant majority disapprove of how he handled the debate?

Solutions

Expert Solution

Solution :

Given that,

n = 1033

x = 473

= x / n = 473 / 1033 = 0.458

1 - = 1 - 0.458 = 0.542

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 * (((0.458 * 0.542) /1033 )

= 0.030

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.458 - 0.030 < p < 0.458 + 0.030

0.428 < p < 0.488

The 95% confidence interval for the population proportion p is : ( 0.428 , 0.488 )


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