Question

in march 2003, the pew research group surveyed adult americans and asked, \do you believe the united states made the right or wrong decision to use military force in iraq?\ in a 1,282 adult portion of americans from the survey, 919 stated the united states made the right decision. the pew research group then asked the same question to a group of adult americans in august 2010. in a 1,282 adult portion of that survey, 535 believed the united states made the right decision. complete parts (a) and (b). click here to view the standard normal distribution table (page 1). click here to view the standard normal distribution table (page 2). (a) in the survey question, the choices
ight\ and \wrong\ were randomly rotated. why? choose the correct answer. a. to remove any potential nonsampling error due to the respondent hearing the word
ight\ or \wrong\ first. b. to reduce the possibility of a type ii error where the null hypothesis is not rejected even though the alternative hypothesis is true. c. to remove any potential sampling error because the randomization is introduced naturally by the survey question. d. to reduce the possibility of a type i error where the null hypothesis is rejected even though the null hypothesis is true. (b) construct and interpret a 90% confidence interval for the difference between the two population proportions, ( p_{2003}-p_{2010} ). the lower bound of a 90% confidence interval is (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportions

For 2003: $n_1 = 1282$, $x_1=919$, so $\hat{p}_1=\frac{x_1}{n_1}=\frac{919}{1282}\approx0.717$.
For 2010: $n_2 = 1282$, $x_2 = 535$, so $\hat{p}_2=\frac{x_2}{n_2}=\frac{535}{1282}\approx0.417$.

Step2: Calculate the difference in sample proportions

$\hat{p}_1-\hat{p}_2=0.717 - 0.417=0.3$.

Step3: Calculate the standard error

$SE=\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=\sqrt{\frac{0.717\times(1 - 0.717)}{1282}+\frac{0.417\times(1 - 0.417)}{1282}}\approx0.027$.

Step4: Find the critical value

For a 90% confidence interval, the critical value $z_{\alpha/2}=z_{0.05}=1.645$ (from the standard - normal distribution table).

Step5: Calculate the lower bound

Lower bound $=(\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\times SE=0.3-1.645\times0.027\approx0.256$.

Answer:

The lower bound of a 90% confidence interval is $0.256$.