1. The best point estimate for p is the sample proportion. Answer: A
2. I agree with your choice.
SAMPLE PROPORTION = .0.388630
SAMPLE SIZE = 159
UPPER LIMIT = .464240
LOWER LIMIT = .331153
What point estimate for the proportion of students prefer foreign automobiles?
A) .0.388630
B) .464240
C) .331153
D) .0.61137
I am having trouble answering this problem? I don't know where to begin.
2. A marketing research company is estimating which of two soft drinks, Mountain Dew or Coke, college students prefer. A random sample of 88 college students produced the following 95% confidence interval for the proportion of college students who prefer Mountain Dew: (0.071, 0.093)
Is the sample size large enough for the interval to be valid?
A) No, the sample size is too small.
B) Yes, the sample was randomly selected from an approximately normal population.
C) No, since mathml equation the sample size is not large enough.
D) Yes, since we know that there are more than 880 college students.
I would say D is the answer because of the population size. Please check my answer. Thx!
2. I agree with your choice.
To find the point estimate for the proportion of students who prefer foreign automobiles, you can subtract the sample proportion from 1.
Point estimate for the proportion of students who prefer foreign automobiles = 1 - 0.388630 = 0.61137
Therefore, the correct answer is D) 0.61137.
2. In this scenario, a random sample of 88 college students was taken to estimate which soft drink, Mountain Dew or Coke, college students prefer. The calculated 95% confidence interval for the proportion of college students who prefer Mountain Dew is (0.071, 0.093).
To determine if the sample size is large enough for the interval to be valid, you need to consider the requirements for a valid confidence interval. For proportions, a general rule is that both np and n(1-p) should be greater than 10, where n is the sample size and p is the proportion. In this case, the sample size is 88, and the midpoint of the confidence interval is (0.071+0.093)/2 = 0.082.
np = 88 * 0.082 = 7.216, and n(1-p) = 88 * (1 - 0.082) = 80.544
Both np and n(1-p) are greater than 10, so the sample size is indeed large enough for the interval to be valid.
Therefore, the correct answer is B) Yes, the sample was randomly selected from an approximately normal population.
Point Estimate for Foreign Automobiles = 1 - Point Estimate for American Automobiles
= 1 - 0.388630
= 0.611370
Therefore, the correct answer is D) 0.61137.
2. The validity of a confidence interval depends on the sample size and the underlying population distribution. In this case, we are given that the sample size is 88 and the confidence interval is (0.071, 0.093). To determine if the sample size is large enough for the interval to be valid, we need to assess if the sample was randomly selected from an approximately normal population.
It was not mentioned whether the population follows an approximately normal distribution. However, the sample size of 88 is generally considered large enough for the central limit theorem to apply, which allows us to assume that the sampling distribution of sample proportions is approximately normal. Therefore, the correct answer is B) Yes, the sample was randomly selected from an approximately normal population.
Note: The answer choice D) Yes, since we know that there are more than 880 college students is irrelevant in this context, as it does not directly address the validity of the confidence interval.