2. Assume that the population proportion of adults having a college degree is 0.35. A random sample of 200 adults is to be selected to test this claim.

A) What is the shape, mean, and standard deviation of the sampling distribution of the sample proportion for samples of size 200?
B) What is the probability that the sample proportion will fall within 2% of the population proportion?
C) Answer part B if the sample size were increased to 500 adults. How does the answer to part C differ from part B, and why?

A) The shape of the sampling distribution of the sample proportion for samples of size 200 will be approximately normal, thanks to the Central Limit Theorem.

The mean of the sampling distribution is equal to the population proportion, which is 0.35 in this case.

The standard deviation of the sampling distribution, also known as the standard error, can be calculated using the formula:

Standard Deviation (σ) = √[(p * (1 - p)) / n]

Where:
- p is the population proportion (0.35)
- n is the sample size (200)

Substituting the values, we get:
Standard Deviation (σ) = √[(0.35 * (1 - 0.35)) / 200]
= √[(0.35 * 0.65) / 200]
= √(0.2275 / 200)
≈ 0.02392

Therefore, the standard deviation (σ) of the sampling distribution is approximately 0.02392.

B) To calculate the probability that the sample proportion falls within 2% of the population proportion, we need to find the area under the normal curve between two values: 0.35 - 0.02 = 0.33 and 0.35 + 0.02 = 0.37.

First, we need to standardize these values using the formula:

Z = (x - μ) / σ

Where:
- Z is the z-score
- x is the value we want to standardize
- μ is the mean of the sampling distribution (0.35)
- σ is the standard deviation of the sampling distribution (0.02392)

Calculating the z-scores:
Z1 = (0.33 - 0.35) / 0.02392 ≈ -0.835
Z2 = (0.37 - 0.35) / 0.02392 ≈ 0.835

Next, we need to find the area under the normal curve between these two z-scores. We can use a standard normal distribution table or a calculator. Assuming a standard normal distribution table, we find the area to be approximately 0.408

Therefore, the probability that the sample proportion will fall within 2% of the population proportion is approximately 0.408.

C) If the sample size were increased to 500 adults, the standard deviation of the sampling distribution (σ) would decrease because of the larger sample size. The formula for the standard deviation of the sampling distribution remains the same:

Standard Deviation (σ) = √[(p * (1 - p)) / n]

However, the sample size (n) would be 500 instead of 200. Substituting the values, we would get a smaller standard deviation than in part A.

With a smaller standard deviation, the probability of the sample proportion falling within 2% of the population proportion would increase. Although the exact probability calculation would differ, we can expect it to be higher than 0.408 since the larger sample size reduces the variability.

A) To determine the shape, mean, and standard deviation of the sampling distribution of the sample proportion, we need to use the formulas:

Shape: According to the Central Limit Theorem, for large sample sizes, the sampling distribution of the sample proportion follows an approximately normal distribution, regardless of the shape of the population distribution.

Mean: The mean of the sampling distribution of the sample proportion is equal to the population proportion, which is 0.35 in this case.

Standard Deviation: The standard deviation of the sampling distribution of the sample proportion (also known as the standard error) can be calculated using the formula:

Standard Deviation (Standard Error) = √((p * (1 - p)) / n)

where p is the population proportion (0.35) and n is the sample size (200).

So, the shape of the sampling distribution is approximately normal, the mean is 0.35, and the standard deviation (or standard error) is √((0.35 * (1 - 0.35)) / 200).

B) To calculate the probability that the sample proportion will fall within 2% of the population proportion, we need to find the range of proportions within 2% of 0.35.

To do this, we first determine the upper and lower bounds of the range:
Upper Bound = 0.35 + 0.02 = 0.37
Lower Bound = 0.35 - 0.02 = 0.33

Next, we can calculate the z-scores for both bounds using the formula:
z = (x - μ) / σ

For the upper bound:
z_upper = (0.37 - 0.35) / (√((0.35 * (1 - 0.35)) / 200))

For the lower bound:
z_lower = (0.33 - 0.35) / (√((0.35 * (1 - 0.35)) / 200))

Once we have the z-scores, we can find the corresponding probabilities from a standard normal distribution table or using a calculator.

The probability that the sample proportion will fall within 2% of the population proportion is equal to the cumulative probability between the obtained z-scores.

C) If the sample size were increased to 500 adults, the answer to part C would differ from part B because the standard deviation (standard error) is inversely proportional to the square root of the sample size.

Since the standard deviation (standard error) is smaller for a larger sample size, the sampling distribution becomes narrower. This means that the probability of the sample proportion falling within 2% of the population proportion would increase.

To find the new probability for part C, you would need to recalculate the standard deviation using the new sample size of 500 and follow the same steps as in part B to find the range and calculate the z-scores.

I'll give you a few hints.

Use the approximation to the binomial distribution.

Your values are the following:
p = .35, q = 1 - p = .65, n = 200

Find mean and standard deviation.
mean = np = (200)(.35) = ?
sd = √npq = √(200)(.35)(.65) = ?

Determine two z-scores.
z = -.02/(sd/√n)
z = .02/(sd/√n)

Once you have the z-scores, you should be able to determine the probability between the two z-scores using a z-table.

For part C), just increase the sample size and recalculate.

I hope this will help get you started.

mean 114

sd=8.608