Question 1

The heights of young American women,in inches,are Normally distributed with a mean of $\mu$and a standard deviation of $\sigma$ = 2.4.The margin of error for a 99% confidence interval is desired to be ±1 inch.What should be the sample size?
A)7
B)22
C)38
D)39

Accepted Answer

The formula for the margin of error (E) in a confidence interval for a mean is E = Z*(σ/√n), where Z is the Z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size. For a 99% confidence level, the Z-score is approximately 2.576. Given E = 1 inch and σ = 2.4, we solve for n: 1 = 2.576*(2.4/√n), leading to n ≈ 38.22. Since sample size must be a whole number, we round up to the nearest whole number, which is 39.

Question 2

The heights of a simple random sample of 400 male high school sophomores in a Midwestern state are measured.The sample mean is  = 66.2 inches.Suppose that the heights of male high school sophomores follow a Normal distribution with a standard deviation of $\sigma$= 4.1 inches.Suppose the heights of a simple random sample of 100 male sophomores were measured rather than 400.Which of the following statements is true?
A)The margin of error for the 95% confidence interval would increase.
B)The margin of error for the 95% confidence interval would decrease.
C)The margin of error for the 95% confidence interval would stay the same because the level of confidence has not changed.
D)The standard deviation $\sigma$ would decrease.

Accepted Answer

The margin of error for a confidence interval is directly proportional to the sample size. As the sample size decreases from 400 to 100, the margin of error will increase. Therefore, choice A is correct. The level of confidence does not affect the margin of error in this case, so choice C is incorrect. The standard deviation is not affected by the sample size, so choice D is also incorrect.

Question 3

To assess the accuracy of a laboratory scale,a standard weight that is known to weigh exactly 1 gram is repeatedly weighed a total of n times and the mean  is computed.Suppose the scale readings are Normally distributed with an unknown mean of$\mu$ and a standard deviation of $\sigma$= 0.01 grams.How large should n be so that a 95% confidence interval for $\mu$has a margin of error no larger than ± 0.0001?
A)n = 100
B)n = 196
C)n = 10000
D)n = 38416

Accepted Answer

D

Question 4

Battery packs in radio-controlled racing cars need to be able to last pretty long.The distribution of the lifetimes of battery packs made by Lectric Co.is slightly left skewed.Assume that the standard deviation of the lifetime distribution is $\sigma$= 2.5 hours.A simple random sample of 75 battery packs results in a mean of  = 29.6 hours.What is a 90% confidence interval for $\mu$,the true average lifetime of the battery packs made by Lectric Co.?
A)(29.13,30.07)
B)(29.03,30.17)
C)(28.86,30.34)
D)The confidence interval cannot be calculated because the population distribution is not Normal.

Accepted Answer

The answer of Battery packs in radio-controlled racing cars need...

Question 5

The breaking strength of yarn used in the manufacture of woven carpet material is Normally distributed with$\sigma$= 2.4 psi.A random sample of 16 specimens of yarn from a production run was measured for breaking strength,and based on the mean of the sample  ,a confidence interval was found to be (128.7,131.3).If we wanted to have a greater degree of confidence in the calculated interval,what would have to happen to its length?
A)The interval would be shorter.
B)The margin of error would decrease.
C)The confidence level would have a smaller value.
D)The sample size would go up.
E)The interval would increase in length.

Accepted Answer

The answer of The breaking strength of yarn used in...

Question 6

An agricultural researcher plants 25 plots with a new variety of yellow corn.Assume that the yield per acre for the new variety of yellow corn follows a Normal distribution with an unknown mean of $\mu$ and a standard deviation of $\sigma$ = 10 bushels per acre.If the average yield for these 25 plots is  = 150 bushels per acre,what is a 90% confidence interval for $\mu$?
A)150 ± 0.784
B)150 ± 2.00
C)150 ± 3.29
D)150 ± 3.92

Accepted Answer

We use the formula for the confidence interval for the mean:

CI = x̄ ± z* (σ/√n)

where x̄ = 150 (sample mean), σ = 10 (population standard deviation), n = 25 (sample size), and z* is the critical value for a 90% confidence level, which is 1.645 (from a standard normal distribution table).

Substituting the values, we get:

CI = 150 ± 1.645 * (10/√25) = 150 ± 3.29

Therefore, the 90% confidence interval for the population mean yield is 150 ± 3.29 bushels per acre. The closest answer choice is (C).

Question 7

Hoping to attract more shoppers,a city builds a new public parking garage downtown.The city plans to pay for the structure through parking fees.The consultant,who advised the city on this project,randomly selected 44 weekdays.Daily fees collected averaged $126.Based on data from other parking structures,the consultant will assume parking fees at this parking garage are Normally distributed with a standard deviation of $15. If a 95% confidence interval was constructed,how would the margin of error compare to one used to create a 90% confidence interval?
A)The margin of error for the 95% confidence interval would be smaller.
B)The margin of error for the 95% confidence interval would be the same.
C)The margin of error for the 95% confidence interval would be larger.
D)This cannot be determined from the information given.

Accepted Answer

As the confidence level increases, the margin of error also increases. Therefore, the margin of error for a 95% confidence interval will be larger than that of a 90% confidence interval.

Question 8

An agricultural researcher plants 25 plots with a new variety of yellow corn.Assume that the yield per acre for the new variety of yellow corn follows a Normal distribution with an unknown mean of$\mu$ and a standard deviation of $\sigma$= 10 bushels per acre. Which of the following would produce a confidence interval with a smaller margin of error than the 90% confidence interval?
A)Plant only five plots rather than 25 because five are easier to manage and control.
B)Plant 10 plots rather than 25 because a smaller sample size will result in a smaller margin of error.
C)Plant 100 plots rather than 25 because a larger sample size will result in a smaller margin of error.
D)Compute a 99% confidence interval rather than a 90% confidence interval because a higher confidence level will result in a smaller margin of error.

Accepted Answer

Increasing the sample size will decrease the margin of error, so planting 100 plots instead of 25 will produce a smaller margin of error. This is known as the Law of Large Numbers.

Question 9

Suppose we wish to calculate a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university.The interval is to have a margin of error of $2.Assume that the amount spent on books by freshmen has a Normal distribution with a standard deviation of $\sigma$ = $30.How many observations are required to achieve this margin of error?
A)25
B)608
C)609
D)865

Accepted Answer

The margin of error is given by:
Margin of Error = z-value * (standard deviation / sqrt(sample size))
We are given that the margin of error is $2, and we want a 90% confidence interval. From the Normal distribution, we know that a 90% confidence interval corresponds to a z-value of 1.645.
Plugging in the given values, we get:
2 = 1.645 * (30 / sqrt(n))
Solving for n, we get:
n = (1.645^2 * 30^2) / 2^2 = 608.71
Since we cannot have a fractional number of observations, we must round up to the nearest whole number. Therefore, we need at least n = 609 observations. Therefore, the correct answer is C.

Question 10

Central Middle School has calculated a 95% confidence interval for the mean height ($\mu$)of 11-year-old boys at their school and found it to be 56 ± 2 inches.Which of the following could be the 90% confidence interval based on the same data?
A)56 ± 1
B)56 ± 2
C)56 ± 3
D)Without knowing the sample size,any of the above answers could be the 90% confidence interval.

Accepted Answer

A 90% confidence interval is narrower than a 95% confidence interval because it reflects a lower level of confidence, meaning it requires a smaller margin of error around the mean.

Question 11

The scores on the Wechsler Intelligence Scale for Children (WISC)are thought to be Normally distributed with a standard deviation of $\sigma$= 10.A simple random sample of 25 children is taken,and each is given the WISC.The mean of the 25 scores is  = 104.32.Based on these data,what is a 95% confidence interval for $\mu$?
A)104.32 ± 0.78
B)104.32 ± 3.29
C)104.32 ± 3.92
D)104.32 ± 19.60

Accepted Answer

We can use the formula for a confidence interval for the population mean:

Confidence interval = sample mean ± (t-value)*(standard error)

where the t-value is based on the degrees of freedom (df = n-1) and the desired level of confidence. For a 95% confidence interval with df = 24, the t-value is 2.064.

The standard error is calculated as standard deviation / sqrt(sample size), which gives:

standard error = 10 / sqrt(25) = 2

Plugging in the values, we get:

Confidence interval = 104.32 ± (2.064)*(2) = 104.32 ± 4.13

So the 95% confidence interval is 104.32 ± 3.92 (rounded to two decimal places), which corresponds to choice C.

Question 12

Hoping to attract more shoppers,a city builds a new public parking garage downtown.The city plans to pay for the structure through parking fees.The consultant,who advised the city on this project,randomly selected 44 weekdays.Daily fees collected averaged $126.Based on data from other parking structures,the consultant will assume parking fees at this parking garage are Normally distributed with a standard deviation of $15. What is a 90% confidence interval for the mean daily income this parking garage will generate from the parking fees?
A)($107.31,$144.69)
B)($121.57,$130.43)
C)($122.28,$129.72)
D)($125.44,$126.56)

Accepted Answer

First, we need to calculate the margin of error:
Margin of Error = z * (standard deviation / sqrt(sample size))
We are looking for a 90% confidence interval, so the z-value is 1.645.
Standard deviation is given as $15.
Sample size is 44.

Margin of Error = 1.645 * ($15 / sqrt(44)) ≈ $4.72

Next, we need to calculate the mean of the sample:
Sample mean = $126

Finally, we can construct the confidence interval:
Confidence interval = Sample mean ± Margin of Error
Confidence interval = $126 ± $4.72
Confidence interval ≈ ($122.28, $129.72)

Therefore, the best choice is C as it provides the correct 90% confidence interval for the mean daily income this parking garage will generate from the parking fees.

Question 13

The heights of a simple random sample of 400 male high school sophomores in a Midwestern state are measured.The sample mean is  = 66.2 inches.Suppose that the heights of male high school sophomores follow a Normal distribution with a standard deviation of $\sigma$= 4.1 inches.What is a 95% confidence interval for $\mu$?
A)(58.16,74.24)
B)(59.46,72.94)
C)(65.80,66.60)
D)(65.86,66.54)

Accepted Answer

The answer of The heights of a simple random sample...

Question 14

The breaking strength of yarn used in the manufacture of woven carpet material is Normally distributed with $\sigma$= 2.4 psi.A random sample of 16 specimens of yarn from a production run was measured for breaking strength,and based on the mean of the sample  ,a confidence interval was found to be (128.7,131.3).What is the confidence level,C,of this interval?
A)0.95
B)0.99
C)0.90
D)0.97
E)It can't be determined with the information provided.

Accepted Answer

The answer of The breaking strength of yarn used in...

Question 15

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M)test this year follows a Normal distribution with standard deviation $\sigma$= 100.You read a report that says,"On the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year,a confidence interval for $\mu$ is found to be 512.00 ± 25.76." What was the confidence level used to calculate this confidence interval?
A)90%
B)95%
C)99%
D)99.9%

Accepted Answer

The answer of Suppose that the population of the scores...

Question 16

A nationally distributed college newspaper conducts a survey among students nationwide every year.This year,responses from a simple random sample of 204 college students to the question "About how many CDs do you own?" resulted in a sample mean of  = 72.8.Based on data from previous years,the editors of the newspaper will assume that $\sigma$ = 7.2.Use the information given to obtain a 95% confidence interval for the mean number of CDs owned by all college students.
A)(65.6,80.0)
B)(71.8,73.8)
C)(72.0,73.6)
D)(72.3,73.3)

Accepted Answer

The answer of A nationally distributed college newspaper conducts a...

Question 17

Suppose a 99% confidence interval for the mean weight of high school girls (in pounds)is calculated as (102.3,106.5).If we had measured the weights of each of the girls in kilograms (2.2 pounds = 1 kilogram),then the confidence interval for the mean weight of high school girls in kilograms would have been
A)(104.5,108.7).
B)(46.5,48.4).
C)(225.06,234.3).
D)(100.1,104.3).

Accepted Answer

To convert the confidence interval from pounds to kilograms, divide each bound by 2.2 (since 2.2 pounds = 1 kilogram). Thus, (102.3 / 2.2, 106.5 / 2.2) = (46.5, 48.4).

Question 18

The heights of young American women,in inches,are Normally distributed with a mean of $\mu$ and a standard deviation of $\sigma$= 2.4.A simple random sample of four young American women is selected and their heights measured.The four heights,in inches,are 63,69,62 and 66.Based on these data,what is a 99% confidence interval for $\mu$?
A)65.00 ± 1.55
B)65.00 ± 2.35
C)65.00 ± 3.09
D)65.00 ± 4.07

Accepted Answer

The sample mean is 65.00. For a 99% confidence interval with a sample size of 4, use the t-distribution with 3 degrees of freedom. The critical t-value is approximately 4.032. The margin of error is t * (s/√n) = 4.032 * (2.4/√4) = 3.09.

Question 19

The distribution of GPA scores is known to be left skewed.At a large university,an English professor is interested in learning about the average GPA score of the English majors and minors.A simple random sample of 75 junior and senior English majors and minors results in an average GPA score of 2.97.Assume that the distribution of GPA scores for all English majors and minors at this university is also left skewed with a standard deviation of 0.62.Calculate a 95% confidence interval for the mean GPA of the junior and senior English majors and minors.
A)(2.786,3.154)
B)(2.830,3.110)
C)(2.852,3.088)
D)(2.954,2.986)

Accepted Answer

The confidence interval is calculated using the formula: mean ± (Z * (standard deviation/√n)). Here, the mean is 2.97, the standard deviation is 0.62, n is 75, and Z for 95% confidence is 1.96. This calculation results in the interval (2.830, 3.110).

Question 20

Researchers are studying the yield of a crop in two locations.The researchers are going to compute independent 90% confidence intervals for the mean yield at each location.What is the probability that at least one of the intervals will cover the true mean yields at their location?
A)0.81
B)0.19
C)0.99
D)0.95

Accepted Answer

The probability that one interval misses the true mean yield is 10% (since we are constructing a 90% confidence interval). Therefore, the probability that both intervals miss the true mean yield is 10% * 10% = 1%.

So the probability that at least one of the intervals covers the true mean yield is 1 - 1% = 99%. Thus, the best choice is C.

Quiz 6: Introduction to Inference

The heights of young American women,in inches,are Normally distributed with a mean of $\mu$ and a standard deviation of $\sigma$ = 2.4.The margin of error for a 99% confidence interval is desired to be ±1 inch.What should be the sample size?

Central Middle School has calculated a 95% confidence interval for the mean height ( $\mu$ ) of 11-year-old boys at their school and found it to be 56 ± 2 inches.Which of the following could be the 90% confidence interval based on the same data?

Researchers are studying the yield of a crop in two locations.The researchers are going to compute independent 90% confidence intervals for the mean yield at each location.What is the probability that at least one of the intervals will cover the true mean yields at their location?

Quiz 6: Introduction to Inference

The heights of young American women,in inches,are Normally distributed with a mean of μ\muμ and a standard deviation of σ\sigmaσ = 2.4.The margin of error for a 99% confidence interval is desired to be ±1 inch.What should be the sample size?

Central Middle School has calculated a 95% confidence interval for the mean height ( μ\muμ ) of 11-year-old boys at their school and found it to be 56 ± 2 inches.Which of the following could be the 90% confidence interval based on the same data?

Researchers are studying the yield of a crop in two locations.The researchers are going to compute independent 90% confidence intervals for the mean yield at each location.What is the probability that at least one of the intervals will cover the true mean yields at their location?

The heights of young American women,in inches,are Normally distributed with a mean of $\mu$ and a standard deviation of $\sigma$ = 2.4.The margin of error for a 99% confidence interval is desired to be ±1 inch.What should be the sample size?

Central Middle School has calculated a 95% confidence interval for the mean height ( $\mu$ ) of 11-year-old boys at their school and found it to be 56 ± 2 inches.Which of the following could be the 90% confidence interval based on the same data?