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In​ 2003, an organization surveyed 1 comma 508 adult Americans and asked about a certain​ war, "Do you believe the United States made the right or wrong decision to use military​ force?" Of the 1 comma 508 adult Americans​ surveyed, 1 comma 085 stated the United States made the right decision. In​ 2008, the organization asked the same question of 1 comma 508 adult Americans and found that 575 believed the United States made the right decision. Construct and interpret a​ 90% confidence interval for the difference between the two population​ proportions, p 2003 minus p 2008.

User Orangepill
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Answer:


(0.719-0.381) - 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.310


(0.719-0.381) + 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.366

And the 90% confidence interval would be given (0.310;0.366).

We are confident at 90% that the difference between the two proportions is between
0.310 \leq p_A -p_B \leq 0.366

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".


p_A represent the real population proportion for 2003


\hat p_A =(1085)/(1508)=0.719 represent the estimated proportion for 2003


n_A=1508 is the sample size required for 2003


p_B represent the real population proportion for 2008


\hat p_B =(575)/(1508)=0.381 represent the estimated proportion for 2008


n_B=1508 is the sample size required for 2008


z represent the critical value for the margin of error

The population proportion have the following distribution


p \sim N(p,\sqrt{(p(1-p))/(n)})

The confidence interval for the difference of two proportions would be given by this formula


(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}

For the 90% confidence interval the value of
\alpha=1-0.90=0.1 and
\alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.


z_(\alpha/2)=1.64

And replacing into the confidence interval formula we got:


(0.719-0.381) - 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.310


(0.719-0.381) + 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.366

And the 90% confidence interval would be given (0.310;0.366).

We are confident at 90% that the difference between the two proportions is between
0.310 \leq p_A -p_B \leq 0.366

User Neil Essy
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