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.

For Store A:

Number tasters = 258

Number preferring local brand = 167

Proportion = 167/258 = 0.6472

Therefore, the proportion based on preference in store A is 0.6472. Do the same for the other two stores. 

Step-by-step explanation: