I agree with your confidence interval. Nice work.
Since .80 isn't within the interval, you are correct to say no as the answer to that question.
random variable p should be the percent or proportion of student who drive.
Can someone help me and let me know if what I have solved is correct.
A researcher randomly surveyed 300 high school seniors and determined 225 stated they drive a car to high school. We are interested in the population proportion of seniors who drive a car to high school.
a. Define the random variable X for this problem in words.
x will represent seniors that drive a car to school.
b. Define the random variable P’ for this problem in words.
c. Construct a 90% confidence interval (CI) for the population proportion of high school seniors who claim to drive a car to high school. Round your CI to FOUR decimal places.
I got 0.75-1.645(√(0.75)(1-0.75)/300 = 0.7089
and i got 0.7911 for the +. which the answers i got are 0.7089 and 0.7911
d. Is it reasonable to conclude at least 80% of seniors drive a car to high school?
no since the numbers do not equal 80.
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