Question 1

An airport traffic analyst wants to estimate the proportion of daily takeoffs by small business jets (as opposed to commercial passenger jets or other aircraft) with an error of ± 4 percent with 90 percent confidence. What sample size should the analyst use?
A)385
B)601
C)410
D)423

Accepted Answer

To determine the sample size for estimating a proportion with a specific margin of error and confidence level, the formula n = (Z^2 * p * (1-p)) / E^2 is used. For 90% confidence, Z is approximately 1.645. Assuming p = 0.5 for maximum variability, and E = 0.04, the calculation gives a sample size closest to 423.

Question 2

A company wants to estimate the time its trucks take to drive from city A to cityB. The standard deviation is known to be 12 minutes. What sample size is required in order that the error will not exceed ±2 minutes, with 95 percent confidence?
A)12 observations
B)139 observations
C)36 observations
D)129 observations

Accepted Answer

To estimate the sample size required for a given margin of error with a known standard deviation and confidence level, the formula is $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $Z$ is the Z-score corresponding to the desired confidence level (1.96 for 95%), $\sigma$ is the standard deviation (12 minutes), and $E$ is the margin of error (2 minutes). Plugging in the values: $n = \left(\frac{1.96 \cdot 12}{2}\right)^2 = 138.24$, which rounds up to 139 observations.

Question 3

A random sample of 160 commercial customers of PayMor Lumber revealed that 32 had paid their accounts within a month of billing. Can normality be assumed for the sample proportion?
A)Yes.
B)No.
C)Need more information to say.

Accepted Answer

In this case, we can use the normal distribution to approximate the sampling distribution of the sample proportion because the sample size is large enough (n=160) and the expected number of successes and failures (32 and 128, respectively) are both greater than 10. Therefore, normality can be assumed for the sample proportion.

Question 4

Which statement is incorrect? Explain.
A)If p = .50 and n = 100, the standard error of the sample proportion is .05.
B)In a sample size calculation for estimating π, it is conservative to assume π = .50.
C)If n = 250 and p = .06, we cannot assume normality in a confidence interval for π.

Accepted Answer

The standard error of the sample proportion is calculated using the formula $\sqrt{\frac{p(1-p)}{n}}$. For A) using p = .50 and n = 100, the standard error is indeed $\sqrt{\frac{.50(1-.50)}{100}} = .05$, making A correct. For B), assuming π = .50 is conservative because it maximizes the product p(1-p), which in turn maximizes the standard error, making the sample size calculation conservative. This is because .50 is the point where the variability of the sample proportion is highest. For C), the statement is incorrect because even with n = 250 and p = .06, we can assume normality in a confidence interval for π under the central limit theorem, especially since np and n(1-p) are both greater than 5 (15 and 235, respectively), which is a common rule of thumb for assuming normality of the sampling distribution.

Question 5

Ersatz Beneficial Insurance wants to estimate the cost of damage to cars due to accidents. The standard deviation of the cost is known to be $200. They want to estimate the mean cost using a 95 percent confidence interval within ± $10. What is the minimum sample size n?
A)1083
B)4002
C)1537
D)2301

Accepted Answer

We use the formula for the margin of error of a 95% confidence interval:
ME = 1.96 * sigma / sqrt(n)

We are given that ME <= $10 and sigma = $200. Plugging these values in and solving for n, we get:

10 = 1.96 * 200 / sqrt(n)
sqrt(n) = 1.96 * 200 / 10
n = (1.96 * 200 / 10)^2
n = 1537.6

Since n must be a whole number, we round up to get n = 1537. Therefore, the minimum sample size is 1537, which corresponds to choice C.

Question 6

A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice?
A)± .0447
B)± .0736
C)± .0876
D)± .0894

Accepted Answer

To calculate the width of a 90% confidence interval for the proportion, we use the formula for the margin of error in a proportion: $ME = Z * \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (for 90%, $Z \approx 1.645$), $p$ is the sample proportion, and $n$ is the sample size. Here, $p = \frac{48}{120} = 0.4$, and $n = 120$. Plugging these values into the formula gives $ME = 1.645 * \sqrt{\frac{0.4(0.6)}{120}} \approx 0.0736$. The width of the confidence interval is twice the margin of error, but since the question asks for the margin of error itself (which is half the width), the correct answer is ±0.0736.

Question 7

What is the approximate width of a 90 percent confidence interval for the true population proportion if there are 12 successes in a sample of 25?
A)± .196
B)± .164
C)± .480
D)± .206

Accepted Answer

First, we need to find the sample proportion: 
p-hat = 12/25 = 0.48 
Next, we can use the formula for the margin of error for a proportion: 
ME = z*sqrt((p-hat*(1-p-hat))/n) 
where z is the critical value for the desired confidence level (in this case, 90% confidence corresponds to a z-value of 1.645), p-hat is the sample proportion, and n is the sample size.
ME = 1.645*sqrt((0.48*(1-0.48))/25) = 0.082 
Finally, we can construct the confidence interval: 
CI = p-hat ± ME = 0.48 ± 0.082 = [0.398, 0.562] 
The width of this interval is approximately 0.164, which is answer choice B.

Question 8

In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 51 to 87. Using the Empirical Rule μ ± 3σ to estimate σ, how many students would you need to sample to estimate the true mean score for the class with 90 percent confidence and an error of ±2?
A)About 17 students
B)About 35 students
C)About 188 students
D)About 25 students

Accepted Answer

First, calculate the standard deviation (σ) using the range and the Empirical Rule. The range is 87 - 51 = 36, which covers 6σ (from -3σ to +3σ). Therefore, σ = 36/6 = 6. To estimate the true mean with 90% confidence and an error of ±2, use the formula for the sample size n = (Z*σ/E)^2, where Z is the Z-score for 90% confidence (approximately 1.645) and E is the margin of error (2). Plugging in the values gives n = (1.645*6/2)^2 ≈ 25. Hence, about 25 students would need to be sampled.

Question 9

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95 percent confidence interval for the proportion of women who prefer a female boss is:
A)± .0288
B)± .0105
C)± .0207
D)± .0196

Accepted Answer

To calculate the width of the confidence interval for a proportion, we use the formula:

width = 2 * z * sqrt(p̂(1-p̂)/n)

where z is the z-score for the desired confidence level (1.96 for 95% confidence), p̂ is the sample proportion (81/810 = 0.1), and n is the sample size (810).

Plugging in these values, we get:

width = 2 * 1.96 * sqrt(0.1(1-0.1)/810) ≈ 0.0207

Therefore, the width of the 95% confidence interval for the proportion of women who prefer a female boss is ±0.0207. The closest answer choice is C.

Question 10

A university wants to estimate the average distance that commuter students travel to get to class with an error of ± 3 miles and 90 percent confidence. What sample size would be needed, assuming that travel distances are normally distributed with a range of X = 0 to X = 50 miles, using the Empirical Rule μ ± 3σ to estimate σ.
A)About 28 students
B)About 47 students
C)About 30 students
D)About 21 students

Accepted Answer

The answer of A university wants to estimate the average...

Question 11

A random sample of 160 commercial customers of PayMor Lumber revealed that 32 had paid their accounts within a month of billing. The 95 percent confidence interval for the true proportion of customers who pay within a month would be:
A)0.148 to 0.252
B)0.138 to 0.262
C)0.144 to 0.256
D)0.153 to 0.247

Accepted Answer

The answer of A random sample of 160 commercial customers...

Question 12

A marketing firm is asked to estimate the percentage of existing customers who would purchase a "digital upgrade" to their basic cable TV service. The firm wants 99 percent confidence and an error of ± 5 percent. What is the required sample size (to the next higher integer)?
A)664
B)625
C)801
D)957

Accepted Answer

To achieve a 99% confidence level with a margin of error of ±5%, the formula for sample size is $n = \frac{Z^2 \times p(1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (2.576 for 99%), $p$ is the estimated proportion (0.5 for maximum variability), and $E$ is the margin of error (0.05). Plugging in the values: $n = \frac{2.576^2 \times 0.5(1-0.5)}{0.05^2} \approx 664.3$. Since we round up to the next higher integer, the required sample size is 665, but since that option is not available, the closest provided option is A) 664.

Question 13

Which of the following is not a characteristic of the t distribution?
A)It is a continuous distribution.
B)It has a mean of 0.
C)It is a symmetric distribution.
D)It approaches z as degrees of freedom decrease.

Accepted Answer

The t distribution approaches the z (normal) distribution as degrees of freedom increase, not decrease. This is because with more data, the t distribution's extra uncertainty (due to estimating the population standard deviation from the sample) becomes less significant, making it more similar to the normal distribution.

Question 14

Jolly Blue Giant Health Insurance (JBGHI) is concerned about rising lab test costs and would like to know what proportion of the positive lab tests for prostate cancer are actually proven correct through subsequent biopsy. JBGHI demands a sample large enough to ensure an error of ± 2 percent with 90 percent confidence. What is the necessary sample size?
A)4,148
B)2,401
C)1,692
D)1,604

Accepted Answer

To determine the necessary sample size for a proportion with a specified margin of error and confidence level, the formula n = (Z^2 * p * (1-p)) / E^2 is used. For 90% confidence, Z is approximately 1.645. Assuming p = 0.5 for maximum variability, and E = 0.02, the calculation gives approximately 1,692.

Question 15

Professor York randomly surveyed 240 students at Oxnard University and found that 150 of the students surveyed watch more than 10 hours of television weekly. Develop a 95 percent confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week. The confidence interval is:
A).533 to .717
B).564 to .686
C).552 to .698
D).551 to .739

Accepted Answer

The correct 95% confidence interval can be calculated using the formula for a proportion confidence interval: $CI = \hat{p} \pm Z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z$ is the Z-score for the desired confidence level (1.96 for 95%), and $n$ is the sample size. Here, $\hat{p} = \frac{150}{240} = 0.625$, $n = 240$, and using the formula, the confidence interval is approximately 0.564 to 0.686.

Question 16

A financial institution wishes to estimate the mean balances owed by its credit card customers. The population standard deviation is $300. If a 99 percent confidence interval is used and an interval of ± $75 is desired, how many cardholders should be sampled?
A)3382
B)629
C)87
D)107

Accepted Answer

The formula for the required sample size is:
n = [Z * (σ / E)]^2

where Z is the z-score corresponding to the desired level of confidence (99% in this case), σ is the population standard deviation, and E is the desired margin of error.

Substituting the given values:

n = [2.576 * (300 / 75)]^2
n = 106.89

Rounding up to the nearest integer, the required sample size is 107. Therefore, the best choice is D.

Question 17

Using the conventional polling definition, find the margin of error for a customer satisfaction survey of 225 customers who have recently dined at Applebee's.
A)± 5.0 percent
B)± 4.2 percent
C)± 7.1 percent
D)± 6.5 percent

Accepted Answer

The answer of Using the conventional polling definition, find the...

Question 18

Last week, 108 cars received parking violations in the main university parking lot. Of these, 27 had unpaid parking tickets from a previous violation. Assuming that last week was a random sample of all parking violators, find the 95 percent confidence interval for the percentage of parking violators that have prior unpaid parking tickets.
A)18.1 to 31.9 percent.
B)16.8 to 33.2 percent.
C)15.3 to 34.7 percent.
D)19.5 to 30.5 percent.

Accepted Answer

To find the confidence interval for the percentage of parking violators that have prior unpaid parking tickets, we will use the formula for confidence interval for a proportion:

p̂ ± z*√(p̂(1-p̂)/n)

where p̂ is the sample proportion (number of parking violators with prior unpaid parking tickets divided by the total sample size), z is the critical value for our desired level of confidence (95% corresponds to a z-value of approximately 1.96), and n is the sample size (number of parking violators).

Using the given information, we have:

- Sample proportion: p̂ = 27/108 = 0.25 
- Sample size: n = 108 
- Critical value: z = 1.96

Plugging these values into the formula, we get:

0.25 ± 1.96*√(0.25(1-0.25)/108)

Simplifying this expression gives us the 95% confidence interval:

0.168 to 0.332

Converting these decimal values to percentages, we get:

16.8% to 33.2%

Therefore, the best choice is B: 16.8 to 33.2 percent.

Question 19

What is the approximate width of an 80 percent confidence interval for the true population proportion if there are 12 successes in a sample of 80?
A)± .078
B)± .066
C)± .051
D)± .094

Accepted Answer

To find the width of the 80% confidence, we need to find the margin of error first. 
Margin of Error = Critical Value * Standard Error
The critical value for an 80% confidence interval is 1.28 (using a table or calculator).
The standard error = sqrt[(p-hat*(1-p-hat))/n]
Here, p-hat = 12/80 = 0.15
So, Standard Error = sqrt[(0.15*(1-0.15))/80] = 0.0432
Now, Margin of error = 1.28 * 0.0432 = 0.05530
So, the approximate width of the 80% confidence interval = 2 * Margin of Error = ± 0.1106, which is approximately equal to ± 0.051.
Therefore, the answer is C.

Question 20

In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will be (assuming other things remain the same):
A)about 25 percent of its former width.
B)about two times wider.
C)about 50 percent of its former width.
D)about four times wider.

Accepted Answer

As n increases, the width of the confidence interval decreases as the sample mean becomes a more precise estimate of the population mean. Thus, multiplying n by 4 will result in a width that is approximately 1/4 of its former width (or 25% of its former width). Therefore, the correct answer is choice C.

Quiz 8: Sampling Distributions and Estimation

An airport traffic analyst wants to estimate the proportion of daily takeoffs by small business jets (as opposed to commercial passenger jets or other aircraft) with an error of ± 4 percent with 90 percent confidence. What sample size should the analyst use?

A company wants to estimate the time its trucks take to drive from city A to cityB. The standard deviation is known to be 12 minutes. What sample size is required in order that the error will not exceed ±2 minutes, with 95 percent confidence?

A random sample of 160 commercial customers of PayMor Lumber revealed that 32 had paid their accounts within a month of billing. Can normality be assumed for the sample proportion?

Which statement is incorrect? Explain.

Ersatz Beneficial Insurance wants to estimate the cost of damage to cars due to accidents. The standard deviation of the cost is known to be $200. They want to estimate the mean cost using a 95 percent confidence interval within ± $10. What is the minimum sample size n?

A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice?

What is the approximate width of a 90 percent confidence interval for the true population proportion if there are 12 successes in a sample of 25?

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95 percent confidence interval for the proportion of women who prefer a female boss is:

A random sample of 160 commercial customers of PayMor Lumber revealed that 32 had paid their accounts within a month of billing. The 95 percent confidence interval for the true proportion of customers who pay within a month would be:

A marketing firm is asked to estimate the percentage of existing customers who would purchase a "digital upgrade" to their basic cable TV service. The firm wants 99 percent confidence and an error of ± 5 percent. What is the required sample size (to the next higher integer) ?

Which of the following is not a characteristic of the t distribution?

A financial institution wishes to estimate the mean balances owed by its credit card customers. The population standard deviation is $300. If a 99 percent confidence interval is used and an interval of ± $75 is desired, how many cardholders should be sampled?

Using the conventional polling definition, find the margin of error for a customer satisfaction survey of 225 customers who have recently dined at Applebee's.

What is the approximate width of an 80 percent confidence interval for the true population proportion if there are 12 successes in a sample of 80?

In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will be (assuming other things remain the same) :