Question 1

In November 2018, 50,000 people living in the United States were asked whether they had a computer at home. 43,128 answered yes, they had a computer at home. Provide a 99% confidence interval for the proportion of people living in the United States who have a computer at home.

Accepted Answer

To calculate a 99% confidence interval for the proportion of people with a computer at home, we use the formula for the confidence interval of a proportion: $p \pm z \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion, $z$ is the z-score corresponding to the confidence level (for 99%, $z \approx 2.576$), and $n$ is the sample size. Here, $p = 43128/50000 = 0.86256$, $n = 50000$, and plugging these values into the formula gives us a range, which is closest to the range provided in option B. This calculation accounts for the variability and the confidence level to estimate the true proportion in the population.

Question 2

In 2015, approximately 136,943,000 people visited an emergency room in the United States. Of these, approximately 12,263,000 were subsequently admitted to the hospital. Provide a 99% confidence interval for the proportion of people visiting an emergency room who were subsequently admitted to the hospital.

Accepted Answer

The correct answer is A because the confidence interval is (0.089485, 0.089611).

Question 3

In November 2018, 50,000 people living in the United States were asked whether they had a computer at home. 43,128 answered yes, they had a computer at home. There were 326,766,748 people living in the United States at the time of the survey. Determine an interval with 99% confidence for the number (how many) of people living in the United States who have a computer at home.

Accepted Answer

We can use the formula for a confidence interval for a proportion:

$	ext{CI} = \hat{p} \pm z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $\hat{p}$ is the sample proportion, $z_{\frac{\alpha}{2}}$ is the critical value from the standard normal distribution for the desired confidence level (in this case, 99% corresponds to $\alpha = 0.01$ and $z_{\frac{\alpha}{2}} = 2.576$), and $n$ is the sample size.

Plugging in the values, we get:

$\hat{p} = \frac{43{,}128}{50{,}000} = 0.86256$

$z_{\frac{\alpha}{2}} = 2.576$

$n = 50{,}000$

$	ext{CI} = 0.86256 \pm 2.576\sqrt{\frac{0.86256(1-0.86256)}{50{,}000}} = 0.85619 	ext{ to } 0.86893$

To find the interval for the number of people with a computer at home, we multiply the lower and upper bounds of this interval by the total US population:

Lower: $0.85619 * 326{,}766{,}748 = 280{,}561{,}930$

Upper: $0.86893 * 326{,}766{,}748 = 283{,}143{,}387$

Therefore, the 99% confidence interval for the number of people living in the United States who have a computer at home is 280,561,930 to 283,143,387. The correct choice is C.

Question 4

A random sample of 25,000 ACT test takers had an average score of 21 with a standard deviation of 5. Determine the standard error.

Accepted Answer

The formula for standard error is standard deviation divided by the square root of sample size. So, in this case, the standard error would be 5 / sqrt(25,000) = 0.0316. Therefore, the correct choice is B.

Question 5

A random sample of 100,000 credit sales in a department store showed an average sale of $87.25. From past data, it is known that the standard deviation of the population is $20.00. With a .95 probability, determine the margin of error.

Accepted Answer

The answer of A random sample of 100,000 credit sales...

Question 6

A random sample of 25,000 ACT test takers had an average score of 21 with a standard deviation of 5. What is the 95% confidence interval of the population mean?

Accepted Answer

To find the 95% confidence interval of the population mean, we will use the formula:

CI = x̄ ± z*(σ/√n)

where x̄ = sample mean = 21
σ = population standard deviation = 5
n = sample size = 25,000
z = z-score for the 95% confidence level = 1.96

Substituting the values, we get:

CI = 21 ± 1.96*(5/√25000)
CI = 21 ± 0.062

Therefore, the 95% confidence interval of the population mean is (21-0.062, 21+0.062), which simplifies to (20.938, 21.062).

The closest answer choice to this interval is B: 4.9380 to 5.0620. However, we must remember to convert the sample mean score of 21 back to the ACT scale of 1-36 in order to compare our interval to the score range.

To convert the scale, we can use the formula:

New score = (Old score/36) * 100

Substituting the values, we get:

New score = (21/36) * 100
New score = 58.3

Therefore, the 95% confidence interval of the population mean on the ACT scale is approximately (49.38, 50.62), which falls within the score range of 1-36.

Question 7

Potential sampling error _____as the sample size increases.

Accepted Answer

As the sample size increases, the potential sampling error decreases because a larger sample size provides a better representation of the population, leading to a smaller margin of error.

Question 8

A random sample of 100,000 credit sales in a department store showed an average sale of $87.25. From past data, it is known that the standard deviation of the population is $20.00. What is the 95% confidence interval of the population mean?

Accepted Answer

For a 95% confidence interval, we need to find the z-score for 97.5% (since we want to divide the 5% alpha level into two tails). This corresponds to a z-score of 1.96. 
Using the formula for a confidence interval: 
CI = X̄ ± (z*σ/√n) 
where X̄ is the sample mean, z is the z-score, σ is the population standard deviation, and n is the sample size. 
Plugging in the values: 
CI = 87.25 ± (1.96*20/√100,000) 
Simplifying: 
CI = 87.25 ± 0.12 
So the 95% confidence interval is from $87.13 to $87.37.

Question 9

A random sample of 25,000 ACT test takers had an average score of 21 with a standard deviation of 5. With a .95 probability, determine the margin of error.

Accepted Answer

We can use the formula for margin of error:
Margin of Error = z*(standard deviation/square root of sample size)
For a 95% confidence level, the z-score is 1.96. 
So, Margin of Error = 1.96*(5/sqrt(25000)) = 0.062 or 0.0620 (rounded to 4 decimal places) which is option D.

Question 10

In 2015, approximately 136,943,000 people visited an emergency room in the United States. Of these, approximately 12,263,000 were subsequently admitted to the hospital. Determine an interval with 99% confidence for the number (how many) of people visiting an emergency room who were subsequently admitted to the hospital.

Accepted Answer

We can use the formula for a confidence interval of a binomial proportion:

p ± z*sqrt((p*(1-p))/n)

where p is the sample proportion (number of admitted patients divided by total number of ER visits), n is the sample size (total number of ER visits), and z* is the critical value from the standard normal distribution for the desired level of confidence (0.99 in this case).

Plugging in the values, we get:

p = 12,263,000 / 136,943,000 = 0.0896
n = 136,943,000
z* = 2.576 (from standard normal distribution table for 99% confidence)

p ± z*sqrt((p*(1-p))/n) = 0.0896 ± 2.576*sqrt((0.0896*(1-0.0896))/136,943,000) = 0.0896 ± 0.00026

Therefore, the 99% confidence interval is [0.08934, 0.08986]. Multiplying this by the total number of ER visits, we get:

[0.08934*136,943,000, 0.08986*136,943,000] = [12,263,159, 12,276,558]

Rounding to the nearest integer, we get the interval [12,263,000, 12,276,000].

Only option D contains this interval, thus the best choice is D.

Quiz 8: Interval Estimation