Question

Response bias Does the appearance of the interviewer influence how people respond to a survey question? Ken (white, with blond hair) and Hassan (darker, with Middle Eastern features) conducted an experiment to address this question. They took turns (in a random order) walking up to people on the main street of a small town, identifying themselves as students from a local high school, and asking them, “Do you support President Obama’s decision to launch airstrikes in Iraq?” Of the $50$people Hassan spoke to, $11$said “Yes,” while $21$of the $44$people Ken spoke to said “Yes.” Construct and interpret a $90\%$confidence interval for the difference in the proportion of people like these who would say they support President Obama’s decision when asked by Hassan versus when asked by Ken.

Short answer

The confidence interval is$(-0.4142,-0.1004)$

Step-by-step solution

Given Information

It is given that $x_1=11$

$x_2=21$

$n_1=50$

$n_2=44$

$c=90\%=0.90$

Checking Conditions

The conditions are:

Random: Samples are independent random samples.

Independent: $50$people are surely less than $10\%$of all people.

Normal: Number of success are $11,21$and number of failures are $50-11=39\mathrm{and}44-21=23$which are greater than ten.

All conditions are satisfied.

Calculation

Sample proportion: ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{11}{50}=0.22$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{21}{44}=0.4773$

For confidence level $1-\alpha =0.90$, using table $z$score is $z_{a/2}=1.645$

Confidence Interval is: $\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{a/2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.22-0.4773)-1.645\times \sqrt{\frac{0.22(1-0.22)}{50}+\frac{0.4773(1-0.4773)}{44}}$

$\approx -0.4142$

and $\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.22-0.4773)+1.645\times \sqrt{\frac{0.22(1-0.22)}{50}+\frac{0.4773(1-0.4773)}{44}}$

$\approx -0.1004$

Hence, confidence interval is$(-0.4142,-0.1004)$