I am having a lot of trouble with these questions can anyone help me please.

2. On September 2, 2004, a Dan Jones poll based on a random sample of 408 Salt Lake County residents reported that 26% of the respondents thought that Mayor Nancy Workman should resign. Find a 95% confidence interval for the percent of all residents in Salt Lake County that felt Nancy Workman should resign.
a. 22% to 30%
b. 24% to 28%
c. 70% to 78%
d. 17% to 35%
3. A health center noted that in a simple random sample of 600 patients, 120 were referred to them by the local hospital. An approximate 90% confidence interval for the percentage of health center patients who were referred by the hospital is:
a. 17.31% to 22.69%
b. 13.4% to 26.6%
c. 16.73% to 23.27%
d. 15.10% to 24.90%
4. A candy company is market-testing a new brand of chocolate bar. A simple random sample of 400 people were given the new chocolate bar to try. A total of 285 said they liked the candy. Choose the statement that is FALSE.
a. The sample percentage is 71.25%; the SE for the sample percentage is 2.26%.
b. 66.73% to 75.77% is a 95% confidence interval for the population percentage of people who will like the new chocolate bar.
c. 66.73% to 75.77% is a 95% confidence interval for the sample percentage of people who will like the new chocolate bar.
d. The candy company can be 95% confident that the true percentage of consumers who will like their new chocolate bar is between 66.73% and 75.77%.
5. A box of tickets has an average of 32 with an SD of 6. Two hundred draws are made at random with replacement from this box. Find the approximate chance that the average of the draws will be between 31.3 and 32.7.
a. 8%
b. 16%
c. 90%
d. 50%

2. 95% = mean ± 1.96 SEm

SEm = SD/√n

3. 90% = mean ± 1.645 SEm

5. Z = (score-mean)/SEm

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores. The values for the SEm (1.96, !.645) on previous problems were also obtained from this table.

2. To find the 95% confidence interval, we can use the formula:

Confidence Interval = sample proportion ± margin of error

The sample proportion can be calculated by dividing the number of respondents who thought Mayor Nancy Workman should resign (408 * 0.26) by the total sample size (408).

Sample proportion = (408 * 0.26) / 408 = 0.26

The margin of error can be calculated using the formula:

Margin of Error = Critical value * Standard Error

The critical value for a 95% confidence level is approximately 1.96 (obtained from a Z-table).

The standard error can be calculated using the formula:

Standard Error = sqrt((sample proportion * (1 - sample proportion)) / sample size)

Standard Error = sqrt((0.26 * (1 - 0.26)) / 408) ≈ 0.0224

Margin of Error = 1.96 * 0.0224 ≈ 0.0439

Now we can calculate the confidence interval:

Confidence Interval = 0.26 ± 0.0439

Confidence Interval ≈ (0.2161, 0.3039)

Therefore, the 95% confidence interval for the percent of all residents in Salt Lake County that felt Nancy Workman should resign is approximately (22% to 30%).

Answer: a. 22% to 30%

3. To find the approximate 90% confidence interval, we can use the formula:

Confidence Interval = sample proportion ± margin of error

The sample proportion can be calculated by dividing the number of patients referred by the hospital (120) by the total sample size (600).

Sample proportion = 120 / 600 = 0.2

The margin of error can be calculated using the formula:

Margin of Error = Critical value * Standard Error

The critical value for a 90% confidence level is approximately 1.645 (obtained from a Z-table).

The standard error can be calculated using the formula:

Standard Error = sqrt((sample proportion * (1 - sample proportion)) / sample size)

Standard Error = sqrt((0.2 * (1 - 0.2)) / 600) ≈ 0.0122

Margin of Error = 1.645 * 0.0122 ≈ 0.0201

Now we can calculate the confidence interval:

Confidence Interval = 0.2 ± 0.0201

Confidence Interval ≈ (0.1799, 0.2201)

Therefore, the approximate 90% confidence interval for the percentage of health center patients referred by the hospital is approximately (17.99% to 22.01%).

Answer: a. 17.31% to 22.69%

4. The false statement is: c. 66.73% to 75.77% is a 95% confidence interval for the sample percentage of people who will like the new chocolate bar.

The confidence interval for the sample percentage of people who liked the new chocolate bar can indeed be calculated, but it does not imply that the population percentage will fall within the same interval.

Answer: c. 66.73% to 75.77% is a 95% confidence interval for the sample percentage of people who will like the new chocolate bar.

5. To find the approximate chance that the average of the draws will be between 31.3 and 32.7, we can use the Central Limit Theorem.

First, we calculate the standard deviation of the sample mean:

Standard Deviation of Sample Mean = Standard Deviation of Population / sqrt(sample size)

Standard Deviation of Sample Mean = 6 / sqrt(200) ≈ 0.4243

Next, we standardize the values (31.3 and 32.7) by subtracting the population mean (32) and dividing by the standard deviation of the sample mean:

Z-score (31.3) = (31.3 - 32) / 0.4243 ≈ -1.85
Z-score (32.7) = (32.7 - 32) / 0.4243 ≈ 1.85

We can then look up the Z-values in a Z-table to find the corresponding probabilities:

Probability (Z ≤ -1.85) ≈ 0.0322
Probability (Z ≤ 1.85) ≈ 0.9678

To find the approximate chance that the average of the draws will be between 31.3 and 32.7, we subtract the probability (Z ≤ -1.85) from the probability (Z ≤ 1.85):

Probability (-1.85 ≤ Z ≤ 1.85) ≈ 0.9678 - 0.0322 ≈ 0.9356

Therefore, the approximate chance that the average of the draws will be between 31.3 and 32.7 is approximately 93.56%.

Answer: c. 90%

I can help you with these questions. Let's go through them one by one and explain how to find the answers.

2. To find a confidence interval, we need to calculate the margin of error and then add/subtract it from the sample proportion.

First, calculate the margin of error using the sample proportion (26%) and the sample size (408):
Margin of Error = Z * sqrt(p * (1-p) / n)
Z is the Z-score corresponding to the desired confidence level. For a 95% confidence level, Z is approximately 1.96.

p is the sample proportion (26% or 0.26) and n is the sample size (408).

Once you've calculated the margin of error, subtract it from 26% and add it to 26% to get the lower and upper limits of the confidence interval.

The correct answer is (b) 24% to 28%.

3. Similar to question 2, we need to calculate the margin of error and then add/subtract it from the sample proportion.

Margin of Error = Z * sqrt(p * (1-p) / n)

Z is the Z-score corresponding to the desired confidence level. For a 90% confidence level, Z is approximately 1.645.

p is the sample proportion (120/600 or 0.2) and n is the sample size (600).

Once you've calculated the margin of error, subtract it from 20% and add it to 20% to get the lower and upper limits of the confidence interval.

The correct answer is (c) 16.73% to 23.27%.

4. To determine the false statement, let's evaluate each statement provided.

a. The statement is true. The sample percentage is calculated as 285/400 = 0.7125 or 71.25%. The standard error (SE) can be calculated as sqrt((p * (1-p)) / n), where p is the sample proportion and n is the sample size.

b. The statement is true. A 95% confidence interval can be calculated using the sample proportion and the standard error.

c. The statement is true. The given interval (66.73% to 75.77%) is a 95% confidence interval for the sample percentage.

d. The statement is false. We can only make a statement about the confidence interval for the sample percentage, not the true population percentage. The confidence interval doesn't guarantee the true percentage for the population.

Therefore, the correct answer is (d) The candy company can be 95% confident that the true percentage of consumers who will like their new chocolate bar is between 66.73% and 75.77%.

5. To find the approximate chance, we can use the Central Limit Theorem. According to the Central Limit Theorem, when the sample size is large enough, the sample mean will be normally distributed, regardless of the distribution of the population.

In this case, the sample mean is normally distributed with a mean of 32 and a standard deviation of 6/sqrt(200) (since it is the average of 200 draws).

Now, we can calculate the Z-scores for the lower and upper limits:
Z(lower) = (31.3 - 32) / (6/sqrt(200))
Z(upper) = (32.7 - 32) / (6/sqrt(200))

Then, we can use a standard normal distribution table or a calculator to find the respective probabilities.

The correct answer is (c) 90%.