Question 1

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 2

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

The sample size is large enough (n=160) for the Central Limit Theorem to apply, which means that the distribution of the sample proportion will be approximately normal. Therefore, normality can be assumed for the sample proportion.

Question 3

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

The answer of An airport traffic analyst wants to estimate...

Question 4

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

The answer of Jolly Blue Giant Health Insurance (JBGHI)is concerned...

Question 5

A marketing firm is asked to estimate the percent 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

The sample size is calculated using the formula: n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level, p is the estimated proportion of customers who would purchase the upgrade, q is the complement of p (1 - p), and e is the desired margin of error. Plugging in the given values, we get: n = (2.576^2 * 0.5 * 0.5) / 0.05^2 = 663.96. Rounding up to the next higher integer, the required sample size is 664.

Question 6

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 margin...

Question 7

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 8

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

We can use the formula for minimum sample size, n = (Z(alpha/2)*sigma/E)^2, where Z(alpha/2) is the z-score for the desired level of confidence (99% in this case), sigma is the population standard deviation ($300), and E is the desired margin of error ($75).
Plugging in the values, we get:
n = (Z(alpha/2)*sigma/E)^2
n = (Z(0.005)*300/75)^2
n = (2.576*300/75)^2
n = 107.38 (rounded up to the nearest whole number)
Therefore, we need to sample at least 107 credit card customers. 
The closest option is D, which is 107.

Question 9

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

When p is small (in this case, p = 0.06), we use the rule n × p ≥ 10 and n × (1 − p) ≥ 10 to determine if normality can be assumed. In this case, we have n × p = 15 and n × (1 − p) = 235, so we can assume normality. Therefore, statement C is incorrect.

The best choice is:

Answer: 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

The sample proportion is in the middle of the confidence interval for the population proportion:
A) in any sample.
B) only if the samples are large.
C) only if π is not too far from .50.

Accepted Answer

The answer of The sample proportion is in the middle...

Question 12

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 13

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 need to use the formula for the margin of error:
Margin of error (E) = z* (sigma/sqrt(n))
We know that E = $10, sigma = $200, and z* for a 95% confidence interval is 1.96.
Substituting the values in the formula, we get:
10 = 1.96 * (200/sqrt(n))
Solving for n, we get:
n = (1.96 * 200 / 10)^2 = 1536.64
Rounding up to the nearest integer, the minimum sample size required would be 1537.
Therefore, the answer is option C.

Question 14

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

Increasing the sample size n to 4n reduces the standard error by a factor of 1/2 (since standard error is inversely proportional to the square root of n), thus reducing the width of the confidence interval to about 50% of its former width.

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.How many additional students would Professor York have to sample to estimate the proportion of all Oxnard University students who watch more than 10 hours of television each week within ± 3 percent with 99 percent confidence?
A) 761
B) 1001
C) 1489
D) 1728

Accepted Answer

We can use the following formula to find the sample size needed:

n = (Z-score)^2 * p * (1-p) / E^2

Where:

- Z-score for 99% confidence level is 2.576
- p is the proportion of students who watch more than 10 hours of TV in the population (unknown)
- E is the desired margin of error, which is 3% or 0.03

We can estimate the value of p as the proportion observed in the sample:

p-hat = 150/240 = 0.625

Now we can plug in the values and solve for n:

n = (2.576)^2 * 0.625 * (1-0.625) / (0.03)^2 = 1488.9

Rounding up, we get a sample size of 1489. Since Professor York has already surveyed 240 students, she would need to sample an additional:

1489 - 240 = 1249

students to meet the desired level of precision. Therefore, the correct answer is C (1489).

Question 16

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 approximate width of the 80 percent confidence interval, we need to use the formula:
Margin of Error = z* * sqrt(p-hat(1-p-hat)/n)
where z* is the z-score for 80% confidence level, p-hat is the sample proportion, and n is the sample size.

To find the z-score for 80% confidence level, we can use a standard normal distribution table or calculator. The z-score for 80% confidence level is 1.28 (rounded to two decimal places).

p-hat = 12/80 = 0.15

n = 80

Now, we can plug in these values into the formula:

Margin of Error = 1.28 * sqrt(0.15*(1-0.15)/80) ≈ 0.051

Therefore, the approximate width of the 80 percent confidence interval is ±0.051. Hence, the choice is C.

Question 17

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 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

We can use the formula for a confidence interval for a proportion:
CI = p̂ ± z*sqrt(p̂(1-p̂)/n)
where p̂ is the sample proportion, z* is the z-score for the desired confidence level (in this case, 1.96 for 95% confidence) and n is the sample size (108).

The sample proportion with prior unpaid parking tickets is 27/108 = 0.25.
Plugging in the values, we get:
CI = 0.25 ± 1.96*sqrt(0.25(1-0.25)/108)
CI = 0.25 ± 0.074
CI = (0.176, 0.324)

Converting to percentages, we get:
CI = (17.6%, 32.4%)

The best choice is B, 16.8 to 33.2 percent, which is the range that includes the entire confidence interval.

Question 19

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, we first need to find the standard error (SE) of the proportion:
SE = sqrt[(p-hat)(1-p-hat)/n]
where p-hat is the sample proportion (81/810 = 0.1) and n is the sample size (810).

SE = sqrt[(0.1)(0.9)/810] = 0.013

The margin of error (ME) for a 95% confidence interval is 1.96 times the standard error:
ME = 1.96 x SE = 0.0255

Therefore, the width of the 95% confidence interval is twice the margin of error:
Width = 2 x ME = 0.051

So the answer is C) ± .0207, which is the closest value to 0.051.

Question 20

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...

Quiz 8: Sampling Distributions and Estimation

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?

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 marketing firm is asked to estimate the percent 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) ?

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.

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?

Which statement is incorrect? Explain.

The sample proportion is in the middle of the confidence interval for the population proportion:

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?

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?

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

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?

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: