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.
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.
To determine if what you have solved is correct, let's go through the steps of constructing a confidence interval for the proportion.
a. The random variable X represents the number of high school seniors who drive a car to school. So, if x is the number of high school seniors who drive a car, then X is the random variable representing this quantity.
b. The random variable P' (pronounced "P prime") represents the sample proportion of high school seniors who drive a car to school. In this case, P' would be the ratio of x to the total sample size, i.e., P' = x/300.
c. To construct a 90% confidence interval for the population proportion of high school seniors who drive a car to school, you can use the formula:
CI = P' ± Z * sqrt((P' * (1 - P')) / n),
where Z is the critical value corresponding to the desired confidence level (in this case, for a 90% confidence level, Z ≈ 1.645), P' is the sample proportion, (1 - P') is the complement of the sample proportion, and n is the sample size.
Let's calculate the confidence interval using your values:
P' = 225/300 = 0.75
Z = 1.645
n = 300
CI = 0.75 ± 1.645 * sqrt((0.75 * (1 - 0.75)) / 300)
= 0.75 ± 1.645 * sqrt(0.1875/300)
= 0.75 ± 1.645 * sqrt(0.000625)
≈ 0.75 ± 1.645 * 0.025
Now, rounding your CI to four decimal places:
Lower bound = 0.75 - (1.645 * 0.025) ≈ 0.7089
Upper bound = 0.75 + (1.645 * 0.025) ≈ 0.7911
So, your calculated confidence interval is approximately 0.7089 to 0.7911.
d. To determine if it is reasonable to conclude that at least 80% of seniors drive a car to high school, you need to check if the value 0.80 falls within the confidence interval. In this case, 0.80 does not fall within the calculated confidence interval of 0.7089 to 0.7911. Therefore, it is not reasonable to conclude that at least 80% of seniors drive a car to high school based on this sample.
Keep in mind that constructing a confidence interval allows us to estimate the population parameter (in this case, the proportion), but it doesn't guarantee that the true population parameter falls within that interval with a certain level of confidence.