Question

Chapter 10 Homework Question 5 (of 5) value 10.00 points A study showed hat 18 of 32 coll phone users with a headset missed their ext, compared with 8 of 32 talking to a passenger Construct a 99 percent confidence interval for the difference in proportions. (Round your answers to 4 decimal places.) The 99 percent confidence interval is tr Hints References eBook& Resources Hint 1 Stat= Exam , Re-docx ^ O Type here to search Stat l' Exam 1 Re-doo 2 3 4 5 6 8

Answer 1

Here

sample proportion of cell phone users with a headset who missed their exit $\hat{p}_1= \frac{18}{32} = 0.5625$

and sample size $n_1 = 32$

and sample proportion of people who missed their exit while talking to a passenger $\hat{p_2}= \frac{8}{32} = 0.25$

and sample size $n_2 = 32$

pooled proportion $\hat{p}= \frac{18+8}{32+32}= 0.40625$

Margin of error (ME) $=z_{0.005}\sqrt{\hat{p}*(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} =(2.57583*0.122783) = 0.316268$

a 99% confidence interval for the difference in proportions

$=((\hat{p_1} - \hat{p_2})-ME, (\hat{p_1} - \hat{p_2})+ME)$

$=((0.5625 - 0.25)- 0.316268, (0.5625 - 0.25)+ 0.316268)$

$=(-0.003768, 0.628768)$