A national opinion poll found that 44% of all American adults agree that parents should be given vouchers good for education at any public or private school of their choice. Suppose that in fact the population proportion who feel this way is p=0.44.

A) Many opinion polls have a "margin of error" of about plus or minus 3%. What is the probability that an SRS of size 300 has a sample proportion p-hat that is within plus or minus 3%(plus or minus .03)of the population proportion p=.44? (show how you would do it using a graphing calculator)

b) Answer the same question for SRSs of sizes 600 and 1200. What is the effect of increasing the size of the sample?

I need the work shown but the answers are:
a) Find P(0.41 ¡Ü p(hat) ¡Ü 0.47)=P(123 ¡Ü X ¡Ü 141. Software gives 0.7309.

b) For n=600, software gives 0.8719. For n=1200, software gives 0.9663. Larger sample sizes are more likely to produce values of p-hat close to the true value of p.

To solve this problem using a graphing calculator, we can use the Normal distribution since the sample size is large enough.

A) To find the probability that the sample proportion, p-hat, is within plus or minus 3% of the population proportion, p = 0.44, with a sample size of 300:

1. Start by calculating the standard deviation of the sample proportion, which is given by:
σ = sqrt((p * (1 - p)) / n)
σ = sqrt((0.44 * (1 - 0.44)) / 300) ≈ 0.025

2. Now, standardize the interval [0.41, 0.47] using the formula:
z = (x - μ) / σ
where x is the value you want P(0.41 ≤ p-hat ≤ 0.47) for, and μ is the population proportion, p.
For this problem, we will use x = 0.41 and x = 0.47.

z1 = (0.41 - 0.44) / 0.025 ≈ -1.2
z2 = (0.47 - 0.44) / 0.025 ≈ 1.2

3. Use the calculator to find P(-1.2 ≤ z ≤ 1.2) using the Normal distribution.
P(-1.2 ≤ z ≤ 1.2) is the probability we're looking for.

B) To answer the same question for sample sizes of 600 and 1200:
Follow the same steps as above, but:
- For n = 600, recalculate the standard deviation with n = 600 in Step 1.
- For n = 1200, recalculate the standard deviation with n = 1200 in Step 1.

Then, calculate P(-1.2 ≤ z ≤ 1.2) for each sample size using the Normal distribution.

The effect of increasing the size of the sample is evident in the results:
- For n = 600, the probability of p-hat being within plus or minus 3% of p increases to 0.8719.
- For n = 1200, the probability further increases to 0.9663.

Larger sample sizes have a higher probability of producing sample proportions (p-hat) close to the true proportion (p).

To solve this problem, we will use the normal distribution and the Central Limit Theorem. The Central Limit Theorem states that for a large enough sample size, the sampling distribution of the sample proportion will be approximately normal.

A) To find the probability that an SRS of size 300 has a sample proportion within plus or minus 3% of the population proportion, we need to find P(0.41 ≤ p(hat) ≤ 0.47). This translates to finding the probability that the sample proportion falls between 0.41 and 0.47.

1. Start by calculating the standard deviation (σ) of the sampling distribution, which is given by the formula:
σ = √(p(1-p)/n)

where p is the population proportion and n is the sample size.

In this case, p = 0.44 and n = 300, so σ = √(0.44(1-0.44)/300) ≈ 0.02574.

2. Next, calculate the z-scores corresponding to the lower and upper bounds. Since the distribution is approximately normal, we can use the z-scores to find the probabilities.

Lower bound z-score:
z1 = (0.41 - p) / σ ≈ (0.41 - 0.44) / 0.02574 ≈ -1.166

Upper bound z-score:
z2 = (0.47 - p) / σ ≈ (0.47 - 0.44) / 0.02574 ≈ 1.166

3. Use a graphing calculator or a standard normal distribution table to find the probability between these z-scores, P(-1.166 ≤ Z ≤ 1.166). This will give you the probability that the sample proportion falls within plus or minus 3% of the population proportion.

Using a graphing calculator, you can find this probability to be approximately 0.7309.

B) Repeat the same steps for sample sizes 600 and 1200.

For n = 600:
σ = √(0.44(1-0.44)/600) ≈ 0.01889
Lower bound z-score: ≈ -1.331
Upper bound z-score: ≈ 1.331
Probability ≈ 0.8719

For n = 1200:
σ = √(0.44(1-0.44)/1200) ≈ 0.01365
Lower bound z-score: ≈ -1.331
Upper bound z-score: ≈ 1.331
Probability ≈ 0.9663

As you can see, as the sample size increases, the probability of obtaining a sample proportion close to the true population proportion (p) increases. This is because a larger sample size reduces the sampling variability, resulting in a more accurate estimate of the population proportion.

No