Asample of 900 college freshmen were randomly selected for a national survey. among the survey participants, 372 students were pursuing liberal arts degrees. the sample proportion is 0.413.what is the margin of error for a 99% confidence interval for this sample? what is the lower endpoint for the 99% confidence interval?

a)

b)

Step-by-step explanation:

1) Data given and notation  

n=900 represent the random sample taken    

X=372 represent the students were pursuing liberal arts degrees

estimated proportion of students were pursuing liberal arts degrees

represent the significance level

z would represent the statistic (variable of interest)    

represent the p value (variable of interest)    

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

The margin of error is given by:

If we replace we have:

And replacing into the confidence interval formula we got:

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."