Answer: a. 0.2788
b) 0.26
0≤ p ≤ 0.298
Step-by-step explanation:
Given : Of the 3,515 vehicles counted, 980 were 18-wheelers.
Here, Sample size : n= 3515
Number of 18-wheelers = 980
Then, the point estimate for the proportion of vehicles traveling Interstate 25 during this time period that are 18-wheelers : [tex]\hat{p}=\dfrac{980}{3515}=0.279[/tex]
Critical z-value for 99% confidence interval ( Using z-value table) :
[tex]z=2.576[/tex]
Now, the 99% confidence interval for the proportion of vehicles on Interstate 25 during this time period that are 18-wheelers will be :
[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
i.e. [tex]0.279\pm (2.576)(0.007565)[/tex]
i.e. [tex]0.279\pm 0.019[/tex]
[tex]\approx0.279\pm0.019=(0.279-0.019,\ 0.279+0.019)\\\\=(0.260,\ 0.298)[/tex]
Hence, a 99% confidence interval for the proportion of vehicles on Interstate 25 during this time period that are 18-wheelers :
0.260≤ p ≤ 0.298