**INSTANT DOWNLOAD WITH ANSWERS**

#### Probability And Statistics For Engineering And The Sciences Devore 8th Edition by Jay L. Devore -Test Bank

**Chapter 6 – Point Estimation**

**COMPLETION**

- The objective of __________ is to select a single number such as , based on sample data, that represents a sensible value (good guess) for the true value of the population parameter, such as .

ANS: point estimation

PTS: 1

- Given four observed values: would result in a point estimate for that is equal to __________.

ANS: 5

PTS: 1

- An estimator that has the properties of __________ and __________ will often be regarded as an accurate estimator.

ANS: unbiasedness, minimum variance

PTS: 1

- A point estimator is said to be an __________ estimator of if for every possible value of .

ANS: unbiased

PTS: 1

- The sample median and any trimmed mean are unbiased estimators of the population mean if the random sample from a population that is __________ and __________.

ANS: continuous, symmetric

PTS: 1

- Among all estimators of parameter that are unbiased, choose the one that has minimum variance. The resulting is called the __________ of .

ANS: minimum variance unbiased estimator (MVUE)

PTS: 1

- The standard error of an estimator is the __________ of .

ANS: standard deviation

PTS: 1

- In your text, two important methods were discussed for obtaining point estimates: the method of __________ and the method of __________.

ANS: moments, maximum likelihood

PTS: 1

- Let be a random sample from a probability mass function or probability density function
*f*(*x*). For*k*= 1,2,3,……, the*k*th population moment is denoted by __________, while the*k*th sample moment is __________.

ANS:

PTS: 1

- Let be a random sample of size
*n*from an exponential distribution with parameter . The moment estimator of = __________.

ANS:

PTS: 1

- Let be the maximum likelihood estimates (mle’s) of the parameters . Then the mle of any function
*h*() of these parameters is the function of the mle’s. This result is known as the __________ principle.

ANS: invariance

PTS: 1

**MULTIPLE CHOICE**

- Which of the following statements are true?

a. | A point estimate of a population parameter is a single number that can be regarded as a sensible value of . |

b. | A point estimate of a population parameter is obtained by selecting a suitable statistic and computing its value from the given sample data. The selected statistic is called the point estimator of . |

c. | The sample mean is a point estimator of the population mean . |

d. | The sample variance is a point estimator of the population variance . |

e. | All of the above statements are true. |

ANS: E PTS: 1

- Which of the following statements are not true?

a. | The symbol is customarily used to denote the estimator of parameter and the point estimate resulting from a given sample. |

b. | The equality is read as “the point estimator of |

c. | The difference between and the parameter is referred to as error of estimation. |

d. | None of the above statements is true. |

ANS: B PTS: 1

- Which of the following statements are not always true?

a. | A point estimator is said to be an unbiased estimator of parameter if for every possible value of . |

b. | If the estimator is not unbiased of parameter , the difference is called the bias of . |

c. | A point estimator is unbiased if its probability sampling distribution is always “centered” at the true value of the parameter , where “centered” here means that the median of the distribution of . |

d. | All of the above statements are true. |

ANS: C PTS: 1

- Which of the following statements are not always true?

a. | It is necessary to know the true value of the parameter to determine whether the estimator is unbiased. |

b. | When X is a binomial random variable with parameters n and p, the sample proportion is an unbiased estimator of p. |

c. | When choosing among several different estimators of parameter , select one that is unbiased. |

d. | All of the above statements are not always true. |

ANS: A PTS: 1

- Which of the following statements are true if is a random sample from a distribution with mean ?

a. | |

b. | |

c. | |

d. | |

e. | All of the above statements are true provided that the sample size n > 30. |

ANS: D PTS: 1

- Which of the following statements are true if is a random sample from a distribution with mean ?

a. | The sample mean is always an unbiased estimator of . |

b. | The sample mean is an unbiased estimator of if the distribution is continuous and symmetric. |

c. | Any trimmed mean is an unbiased estimator of if the distribution is continuous and symmetric. |

d. | None of the above statements are true. |

e. | All of the above statements are true. |

ANS: E PTS: 1

- Which of the following statements are not true?

a. | Maximum likelihood estimators are generally preferable to moment estimators because of certain efficiency properties. |

b. | Maximum likelihood estimators often require significantly more computation than do moment estimators. |

c. | The definition of unbiasedness in general indicates how unbiased estimators can be derived. |

d. | None of the above statements are true. |

e. | All of the above statements are true |

ANS: C PTS: 1

- Which of the following statements are correct?

a. | The first population moment is , while the first sample moment is . |

b. | The moment estimators are obtained by equating the first m sample moments to the corresponding first m population moments, and solving for the unknown parameters . |

c. | The method of maximum likelihood was first introduced by R.A. Fisher, a geneticist and statistician, in the 1920’s. |

d. | All of the above statements are true. |

e. | Only (A) and (B) are true. |

ANS: D PTS: 1

- Which of the following statements are not true?

a. | Maximizing the likelihood function gives the parameter values for which the observed sample is most likely to have been generated—that is, the parameter values that “agree most likely” with the observed data. |

b. | Different principles of estimation may yield different estimators of the unknown parameters. |

c. | The maximum likelihood estimator of the population standard deviation is the sample standard deviation S. |

d. | None of the above statements are true. |

ANS: C PTS: 1

- Which of the following statements are true?

a. | Maximizing the likelihood estimation is the most widely used estimation technique among statisticians. |

b. | Under very general conditions on the joint distribution of the sample, when the sample size n is large, the maximum likelihood estimator of any parameter is approximately unbiased; that is, . |

c. | Under very general conditions on the joint distribution of the sample, when the sample size n is large, the maximum likelihood estimator of any parameter has variance, is nearly as small as small as can be achieved by any estimator. |

d. | In recent years, statisticians have proposed an estimator, called an M-estimator, which is based on a generalization of maximum likelihood estimation. |

e. | All of the above are true statements. |

ANS: E PTS: 1

- The accompanying data describe flexural strength (Mpa) for concrete beams of a certain type was introduced in Example 1.2.

9.2 9.7 8.8 10.7 8.4 8.7 10.7

6.9 8.2 8.3 7.3 9.1 7.8 8.0

8.6 7.8 7.5 8.0 7.3 8.9 10.0

8.8 8.7 12.6 12.3 12.8 11.7

a. | Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used. Hint: |

b. | Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50%, and state which estimator you used. |

c. | Calculate and interpret a point estimate of the population standard deviation Which estimator did you use? Hint: |

d. | Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 11 Mpa. Hint: Think of an observation as a “success” if it exceeds 11. |

e. | Calculate a point estimate of the population coefficient of variation and state which estimator you used. |

ANS: A

. We use the sample mean,

- We use the sample median, (the middle observation when arranged in ascending order).
- We use the sample standard deviation,
- With “success” = observation greater than 11, s = # of successes = 4, and

- We use the sample (std dev) /(mean), or

PTS: 1

**ESSAY**

- Answer the following questions.

- A random sample of 10 houses in Big Rapids, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 108, 161, 123, 94, 130, 152, 127, 114, 143, 104. Let denote the average gas usage during January by all houses in this area. Compute a point estimate of .
- Suppose there are 10,000 houses in Big Rapids that use natural gas for heating. Let denote the total amount of gas used by all of these houses during January. Estimate using the data of part (a0. What estimator did you use in computing your estimate?
- Use the data in part (a) to estimate
*p*, the proportion of all houses that used at least 105 therms. - Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

ANS:

- 8 of 10 houses in the sample used at least 100 therms (the “successes”), so

- The ordered sample values are 94, 104, 108, 114, 123, 127, 130, 143, 152, 161, from which the two middle values are 123 and 127, so

PTS: 1

- Consider a random sample from the pdf

where (this distribution arises in particle physics). Show that is an unbiased estimator of [Hint: First determine

ANS:

PTS: 1

- Let represent a random sample from a Rayleigh distribution with pdf

- It can be shown that Use this fact to construct an unbiased estimator of based on (and use rules of expected value to show that it is unbiased).
- Estimate from the following observations on vibratory stress of a turbine blade under specified conditions:

17.08 10.43 4.79 6.86 13.88

14.43 20.07 9.60 6.71 11.15

ANS:

- Then

is an unbiased estimator for .

PTS: 1

- A random sample of bike helmets manufactured by a certain company is selected. Let = the number among the that are flawed and let = (flawed). Assume that only is observed, rather than the sequence of

- Derive the maximum likelihood estimator of . If = 25 and =5, what is the estimate?
- Is the estimator of part (a) unbiased?
- If = 25 and =5, what is the mle of the probability that none of the next five helmets examined is flawed?

ANS:

- We wish to take the derivative of set it equal to zero and solve for

setting this equal to zero and solving for yields . For n = 25 and x = 5,

- is an unbiased estimator of .

PTS: 1

- Let denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of is

where > -1. A random sample of ten students yields data

- Use the method of moments to obtain an estimator of and then compute the estimate for this data.
- Obtain the maximum likelihood estimator of and then compute the estimate for the given data.

ANS:

- so the moment estimator is the solution to

Since

- so the log likelihood is

Taking and equating to 0 yields

Taking

PTS: 1

- The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

395 379 404 370 392 365 412 418 361 378

- Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.
- Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. (Hint: What is the 95 percentile in terms of ? Now use the invariance principle.)

ANS:

and (this is not s).

- The 95
^{th}percentile is , so the mle of this is (by the invariance principle) is given by

PTS: 1

- Consider a random sample from the shifted exponential pdf

- Obtain the maximum likelihood estimators of
- A random sample of size results in the values 3.12, .65, 2.56, 2.21, 5.45, 3.43, 10.40, 8.94, 17.83, and 1.31, calculate the estimates of

ANS:

- The joint pdf (likelihood function) is

Notice that

and that

Thus likelihood =

Consider maximization wrt . Because the exponent is positive, increasing will increase the likelihood provided that if we make larger than the likelihood drops to 0. This implies that the mle of The log likelihood is now

Equating the derivative wrt to 0 and solving yields

PTS: 1

**Chapter 7 – Statistical Intervals Based on a Single Sample**

**COMPLETION**

- The formula used to construct a 95% confidence interval for the meanof a normal population when the value of the standard deviation is known is given by __________.

ANS:

PTS: 1

- If the random sample is taken from a normal distribution with mean valueand standard deviation , then regardless of the sample size
*n*, the sample mean is distributed with expected value __________ and standard deviation __________.

ANS: normally,,

PTS: 1

- The standard normal random variable has a mean value of __________ and standard deviation of __________.

ANS: 0, 1

PTS: 1

- If a confidence level of 90% is used to construct a confidence interval for the meanof a normal population when the value of the standard deviation is known, the
*z*critical value is __________.

ANS: 1.645

PTS: 1

- If you want to develop a 99% confidence interval for the meanof a normal population, when the standard deviation is known, the confidence level is __________.

ANS: .99

PTS: 1

- If we think of the width of the confidence interval as specifying its precision or accuracy, then the confidence level (or reliability) of the interval is __________ related to its precision.

ANS: inversely

PTS: 1

- The ability of a confidence interval to contain the value of the population mean is described by the __________.

ANS: confidence level

PTS: 1

- Let be a random sample from a population having a mean and standard deviation . Provided that
*n*is large, the Central Limit Theorem (CLT) implies that is __________ distributed.

ANS: approximately normally

PTS: 1

- A random sample of 50 observations produced a mean value of 55 and standard deviation of 6.25. The 95% confidence interval for the population mean is between __________ and __________. (two decimal places)

ANS: 53.27, 56.73

PTS: 1

- The formula used to construct approximately confidence interval for a population proportion
*p*when the sample size*n*is large enough is given by __________, where is the sample proportion, and

ANS:

PTS: 1

- A large-sample lower confidence bound for the population mean __________.

ANS:

PTS: 1

- When is the mean of a random sample of size
*n*(*n*is small) from a normal population with mean, the random variablehas a probability distribution called*t*-distribution with*n*-1 __________.

ANS: degrees of freedom

PTS: 1

- When is the mean of a random sample of size
*n*(*n*is large) from a normal population with mean , the random variable has approximately a __________ distribution with mean value of __________ and standard deviation of __________.

ANS: standard normal, 0,1

PTS: 1

- Let denote the density function curve for a
*t*-distribution with degrees of freedom. As __________, the spread of the corresponding curve decreases.

ANS: increases

PTS: 1

- The
*z*curve is often called the*t*curve with degrees of freedom equal to __________.

ANS:

PTS: 1

- The area under a
*t*-density curve between the critical values is __________.

ANS:

PTS: 1

- Let be a random sample from a normal distribution with mean and variance . Then the random variable has a __________ probability distribution with __________ degrees of freedom.

ANS: chi-squared

PTS: 1

- The chi-squared critical value,, denotes the number on the measurement axis such that __________ of the area under the chi-squared curve with __________ degrees of freedom lies to the __________ of .

ANS: right

PTS: 1

- The 5
^{th}percentile of a chi-squared distribution with 10 degrees of freedom is equal to __________.

ANS: 3.94

PTS: 1

- The 90
^{th}percentile of a chi-squared distribution with 15 degrees of freedom is equal to __________.

ANS: 22.307

PTS: 1

- The area under a chi-squared curve with 10 degrees of freedom, which is captured between the two critical values is __________.

ANS:

PTS: 1

**MULTIPLE CHOICE**

- Which of the following statements are true?

a. | A confidence interval is always calculated by first selecting a confidence level, which is a measure of the degree of reliability of the interval. |

b. | A confidence level of 95% implies that 95% of all samples would give an interval that includes the parameter being estimated, and only 5% of all samples would yield an erroneous interval. |

c. | Information about the precision of an interval estimate is conveyed by the width of the interval. |

d. | The higher the confidence level, the more strongly we believe that the value of the parameter being estimated lies within the interval. |

e. | All of the above statements are true. |

ANS: E PTS: 1

- Which of the following statements are true?

a. | The interval is random, while its width is not random. |

b. | The interval is not random, while its width is random. |

c. | The interval is random, while its width is not random. |

d. | The interval is not random, while its width is random. |

e. | None of the above statements are true. |

ANS: A PTS: 1

- Which of the following statements are not true?

a. | A correct interpretation of a confidence interval for the meanrelies on the long-run frequency interpretation of probability. |

b. | It is correct to write a statement such as |

c. | The probability is .95 that the random interval includes or covers the true value of . |

d. | The interval is a 90% confidence interval for the mean . |

e. | None of the above statements are true. |

ANS: B PTS: 1

- If one wants to develop a 90% confidence interval for the mean of a normal population, when the standard deviation is known, the confidence level is

a. | .10 | c. | .90 |

b. | .45 | d. | 1.645 |

ANS: C PTS: 1

- Which of the following statements are true?

a. | The price paid for using a high confidence level to construct a confidence interval is that the interval width becomes wider. |

b. | The only 100% confidence interval for the mean is . |

c. | If we wish to estimate the mean of a normal population when the value of the standard deviation is known, and be within an amount B with confidence, the formula for determining the necessary sample size n is . |

d. | All of the above statements are true. |

e. | None of the above statements are true. |

ANS: D PTS: 1

- A 99% confidence interval for the mean of a normal population when the standard deviation is known is found to be 98.6 to 118.4. If the confidence level is reduced to .95, the confidence interval for

a. | becomes wider | c. | remains unchanged |

b. | becomes narrower | d. | None of the above answers are correct. |

ANS: B PTS: 1

- If the width of a confidence interval for is too wide when the population standard deviation is known, which one of the following is the best action to reduce the interval width?

a. | Increase the confidence level |

b. | Reduce the population standard deviation |

c. | Increase the population mean |

d. | Increase the sample size n |

e. | None of the above answers are correct. |

ANS: D

Section 7.2

PTS: 1

- Which of the following statements are not true?

a. | Provided that the sample size n is large, the standardized variable is approximately normally distributed, while the variable is not. |

b. | The formula is a large-sample confidence interval for with confidence level approximately . |

c. | Generally speaking, n >40 will be sufficient to justify the use of the formula as a large-sample confidence interval for. |

d. | None of the above statements are true. |

e. | All of the above statements are true. |

ANS: A PTS: 1

- A random sample of 64 observations produced a mean value of 82 and standard deviation of 5.5. The 90% confidence interval for the population meanis between

a. | 81.86 and 82.14 |

b. | 80.65 and 83.35 |

c. | 80.87 and 83.13 |

d. | 81.31 and 82.69 |

e. | None of the above answers are correct. |

ANS: C PTS: 1

- A random sample of 100 observations produced a sample proportion of .25. An approximate 90% confidence interval for the population proportion
*p*is

a. | .248 and .252 |

b. | .179 and .321 |

c. | .423 and .567 |

d. | .246 and .254 |

e. | None of the above answers are correct. |

ANS: B PTS: 1

- Which of the following expressions are true about a large-sample upper confidence bound for the population mean?

a. | |

b. | |

c. | |

d. | |

e. | None of the above statements are true. |

ANS: D PTS: 1

- Suppose that an investigator believes that virtually all values in the population are between 38 and 70. The appropriate sample size for estimating the true population mean within 2 units with 95% confidence level is approximately

a. | 61 |

b. | 62 |

c. | 15 |

d. | 16 |

e. | None of the above answers are correct. |

ANS: B PTS: 1

- A 99% confidence interval for the population mean is determined to be (65.32 to 73.54). If the confidence level is reduced to 90%, the 90% confidence interval for

a. | becomes wider | c. | remains unchanged |

b. | becomes narrower | d. | None of the above answers are correct. |

ANS: B PTS: 1

- In developing a confidence interval for the population mean , a sample of 50 observations was used, and the confidence interval was 15.24 1.20. Had the sample size been 200 instead of 50, the confidence interval would have been

a. | 7.62 1.20 |

b. | 15.24 .30 |

c. | 15.24 .60 |

d. | 3.811.20 |

e. | None of the above answers are correct. |

ANS: C PTS: 1

- Which of the following statements are not true in developing a confidence interval for the population mean

a. | The width of the confidence interval becomes narrower when the sample mean increases. |

b. | The width of the confidence interval becomes wider when the sample mean increases. |

c. | The width of the confidence interval becomes narrower when the sample size n increases. |

d. | All of the above statements are true. |

e. | None of the above statements are true. |

ANS: A PTS: 1

- Which of the following statements are true when is the mean of a random sample of size
*n*from a normal distribution with mean ?

a. | The random variable has approximately a standard normal distribution for large n. |

b. | The random variable has a t-distribution with n-1 degrees of freedom for small n. |

c. | The normal distribution is governed by two parameters, the mean and the standard deviation . |

d. | A t-distribution is governed by only one parameter, called the number of degrees of freedom. |

e. | All of the above answers are true. |

ANS: E PTS: 1

- Which of the following statements are not true if denotes the density function curve for a
*t*-distribution with degrees of freedom?

a. | The t-distribution is governed by only. |

b. | Each curve is bell-shaped and centered around 0. |

c. | Each curve is less spread out than the standard normal z curve. |

d. | As increases, the spread of the corresponding curve decreases. |

e. | None of the above answers are true. |

ANS: C PTS: 1

- Which of the following statements are not true?

a. | The notation is often used to denote the number on the measurement axis for which the area under the t-curve with degrees of freedom to the left of is , where is called a t critical value. |

b. | The number of degrees of freedom for a t– variable is the number of freely determined deviations on which the estimated standard deviation in the denominator of is based. |

c. | A larger value of degrees of freedom implies a t-distribution with smaller spread. |

d. | All of the above statements are true. |

e. | None of the above statements are true. |

ANS: A PTS: 1

- A random sample of size 16 is taken from a normal population with mean . If the sample mean is 75 and the sample standard deviation is 5, then a 95% upper confidence bound for is

a. | 77.664 |

b. | 77.191 |

c. | 72.809 |

d. | 72.336 |

e. | None of the above answers are correct. |

ANS: B PTS: 1

- A random sample of 10 observations was selected from a normal population distribution. The sample mean and sample standard deviations were 20 and 3.2, respectively. A 95% prediction interval for a single observation selected from the same population is

a. | 206.152 |

b. | 204.244 |

c. | 207.962 |

d. | 207.592 |

e. | None of the above answers are correct. |

ANS: D PTS: 1

- Which of the following statements are false about the chi-squared distribution with degrees of freedom?

a. | It is a discrete probability distribution with a single parameter . |

b. | It is positively skewed (long upper tail) |

c. | It becomes more symmetric as increases. |

d. | All of the above statements are true. |

e. | All of the above statements are false. |

ANS: A PTS: 1

- Which of the following statements are true about the percentiles of a chi-squared distribution with 20 degrees of freedom?

a. | The 5^{th} percentile is 31.410 |

b. | The 95^{th} percentile is 10.851 |

c. | The 10^{th} percentile is 12.443 |

d. | The 90^{th} percentile is 37.566 |

e. | All of the above statements are true. |

ANS: C PTS: 1

- The lower limit of a 95% confidence interval for the variance of a normal population using a sample of size
*n*and variance value is given by:

a. | |

b. | |

c. | |

d. | |

e. | None of the above answers are correct. |

ANS: B PTS: 1

- The upper limit of a 95% confidence interval for the variance of a normal population using a sample of size
*n*and variance value is given by:

a. | |

b. | |

c. | |

d. | |

e. | None of the above answers are correct. |

ANS: D PTS: 1

**SHORT ANSWER**

- Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected, and the alcohol content of each bottle is determined. Let denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (8.0, 9.6).

- Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.
- Consider the following statement: There is a 95% chance that is between 8 and 9.6. Is this statement correct? Why or why not?
- Consider the following statement: We can be highly confident that 95% of all bottles of this type of cough syrup have an alcohol content that is between 8.0 and 9.6. Is this statement correct? Why or why not?
- Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding 955 interval ire repeated 100 times, 95 of the resulting intervals will include . Is this statement correct? Why or why not?

ANS:

- A 90% confidence interval will be narrower. Also, the z critical value for a 90% confidence level is 1.645, smaller than the z of 1.96 for the 95% confidence level, thus producing a narrower interval.
- Not a correct statement. Once an interval has been created from a sample, the mean is either enclosed by it, or not. The 95% confidence is in the general procedure, for repeated sampling.
- Not a correct statement. The interval is an estimate for the population mean, not a boundary for population values.
- Not a correct statement. In theory, if the process were repeated an infinite number of times, 95% of the intervals would contain the population mean.

PTS: 1

- A CI is desired for the true average stray-load loss (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with = 3.0.

- Compute a 95% CI for when
*n*= 25 and = 60. - Compute a 95% CI for when
*n*= 100 and = 60. - Compute a 99% CI for when
*n*= 100 and = 60. - Compute an 82% CI for when
*n*= 100 and = 60. - How large must
*n*be if the width of the 99% interval for is to be 1.0?

ANS:

** **

- 82% confidence and the interval is

- = 239.62 so n = 240.

PTS: 1

- By how much must the sample size
*n*be increased if the width of the CI is to be halved? If the sample size is increased by a factor of 25, what effect will this have on the width of the interval? Justify your assertions.

ANS:

If L = and we increase the sample size by a factor of 4, the new length is . Thus halving the length requires n to be increased fourfold. If then , so the length is decreased by a factor of 5.

PTS: 1

- Consider the 1000 95% confidence intervals (CI) for that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of? What is the probability that between 950 and 970 of these intervals contain the corresponding value of? (
*Hint*: Let*Y*= the number among the 1000 intervals that contain. What kind of random variable is*Y*?).

ANS:

Y is a binomial r.v. with n = 1000 and p = .95, so E(Y) = np = 950, the expected number of intervals that capture , and Using the normal approximation to the binomial distribution, P(950 Y 970) = P(949.5 970.5) = P(-.07 Z 2.97) = .9985 – .4721 = .5264.

PTS: 1

- A random sample of 100 lightning flashes in a certain region resulted in a sample average radar echo duration of .81 sec and a sample standard deviation of .34 sec. Calculate a 99% (two-sided) confidence interval for the true average echo duration, and interpret the resulting interval.

ANS:

PTS: 1

- Determine the confidence level for each of the following large-sample one-sided confidence bounds:

- Upper bound:
- Lower bound:
- Upper bound:

ANS:

- , so the confidence level is 80%.
- , so the confidence level is 98%.
- , so the confidence level is 75%

PTS: 1

- It was reported that, in a sample of 507 adult Americans, only 142 correctly described the Bill of Rights as the first ten amendments to the U.S. Constitution. Calculate a (two-sided) confidence interval using a 99% confidence level for the proportion of all U. S. adults that could give a correct description of the Bill of Rights.

ANS:

N = 507, x = # of successes = 142 = .28; the 99% two-sided interval is:

PTS: 1

- The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter. Use the accompanying data on absences for 50 days to derive a large-sample CI for . [
*Hint*: The mean and variance of a Poisson variable both equal , so has approximately a standard normal distribution. Now proceed as in the derivation of the interval for*p*by making a probability statement (with probability 1 – ) and solving the resulting inequalities for.

# Absences |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

Frequency |
1 | 6 | 8 | 10 | 8 | 7 | 5 | 3 | 2 | 1 | 1 |

ANS:

With . The large sample C.I. is then . We calculate then given by:

.

PTS: 1

- Determine the
*t*critical value for a two-sided confidence interval in each of the following situations.

- Confidence level = 95%, df = 12
- Confidence level = 95%, df = 15
- Confidence level = 99%, df = 20
- Confidence level = 99%,
*n*= 8 - Confidence level = 98%, df =25
- Confidence level = 99%,
*n*= 40

ANS:

PTS: 1

- A random sample of
*n*= 8 E-glass fiber test specimens of a certain type yielded a sample mean interfacial shear yield stress of 30.5 and a sample standard deviation of 3.0. Assuming that interfacial shear yield stress is normally distributed, compute a 95% CI for true average stress.

ANS:

d.f. = *n* – 1 = 7, so the critical value for a 95% C.I. is = 2.365. The interval is then

.

PTS: 1

- A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of 8.50 MPa and a sample standard deviation of .80 MPa.

- Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?
- Calculate and interpret a 95% lower prediction bound for the proportional limit stress of a single joint of this type.

ANS:

- A 95% lower confidence bound: With 95% confidence, the value of the true mean proportional limit stress of all such joints lies in the interval (8.11, ). If this interval is calculated for sample after sample, in the long run 955 of these intervals will include the true mean proportional limit stress of all such joints. We must assume that the sample observations were taken from a normally distributed population.

- A 95% lower prediction bound:

If this bound is calculated for sample after sample, in the long run 95% of these bounds will provide a lower bound for the corresponding future values of the proportional limit stress of a single joint of this type.

PTS: 1

- A study of the ability of individuals to walk in a straight line reported that accompanying data on cadence (strides per seconds) for a sample of
*n*– 20 randomly selected healthy men:

.95 .81 .93 .95 .93 .86 1.05 .92 .85 .81

.92 .96 .92 1.00 .78 1.06 1.06 .96 .85 .92

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from MINITAB follows.

*Variable N Mean Median StDev SEMean*

*Cadence 20 0.9255 0.9300 0.0809 0.0181*

* *

- Calculate and interpret a 95% confidence interval for a population mean cadence.
- Calculate and interpret a 95% prediction interval for the cadence of a single individual randomly selected from this population.
- Calculate an interval that includes at least 99% of the cadences in the population distribution using a confidence level of 95%.

ANS:

- A 95% C.I.:
- A 95% P.I.:
- A tolerance interval is requested, with k = 99, confidence level 95%, and n = 20. The tolerance critical value, from
**the table of “tolerance critical values for normal population distributions” available in your text,**is 3.615. The interval is

PTS: 1

- A more extensive tabulation of
*t*critical values than what appears in your text shows that for the*t*distribution with 20 df, the areas to the right of the values .687, .860, and 1.064 are .25, .20, and .15, respectively. What is the confidence level for each of the following three confidence intervals for the mean of a normal population distribution? Which of the three intervals would you recommend be used, and why?

ANS:

The 20 d.f. row of **the table of “critical values for t distribution” available in your text** shows that 1.725 captures upper tail area .05 and 1.325 captures upper tail area .10. The confidence level for each interval is 100(central area)%. For the first interval, central area = 1 – sum of tail areas = 1 – (.25 + .05) = .70, and for the second and third intervals the central areas are 1 – (.20 + .10) =.70 and 1 – (.15 + .15) = 70. Thus each interval has confidence level 70%. The width of the first interval is whereas the widths of the second and third intervals are 2.185 and 2.128 respectively. The third interval, with symmetrically placed critical values, is the shortest, so it should be used. This will always be true for a t interval.

PTS: 1

- Determine the values of the following quantities:

ANS:

PTS: 1

- Determine the following:

- The 90
^{th}percentile of the chi-squared distribution with*=*12. - The 10
^{th}percentile of the chi-squared distribution with = 12. - where is a chi-squared rv with = 22.
- where is a chi-squared rv with = 25

ANS:

** **

- Since 10.982 = and 36.78 = ,
- Since 14.61 = and 37.65 = ,

PTS: 1

- The amount of lateral expansion (mils) was determined for a sample of
*n*= 9 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was*s*= 2.80 mils. Assuming normality, derive a 95% CI for and for .

ANS:

*n* – 1 = 8, = 17.543, = 2.180,

The 95% interval for is then

The 95% interval for is

PTS: 1

- The results of a Wagner turbidity test performed on 15 samples of standard Ottawa testing sand were (in microamperes)

26.9 25.8 24.4 24.1 26.4 25.9 24.0 21.7

24.9 25.9 27.3 26.7 26.9 24.8 27.3

- Is it plausible that this sample was selected from a normal population distribution?
- Calculate an upper confidence bound with confidence level 90% for the population standard deviation of turbidity.

ANS:

- Using a normal probability plot, we ascertain that it is plausible that this sample was taken from a normal population distribution.
- With s = 1.579, n = 15, and = 21.064 the 90% upper confidence bound

For is.

PTS: 1