Statistics for scientists and engineers 3rd edition solutions

Simple random samples generally do not reflect the population exactly. Let A denote an acceptable chip, and U an unacceptable one. This is the smallest possible value. With this process, the probability that a ring meets the specification is Z Let X1 , Let X1 and X2 denote the lengths of the pieces chosen from the population with mean 30 and standard deviation 0.

The tank holds 20 gallons of gas. Let Y1 be the number of miles traveled on the first gallon, let Y2 be the number of miles traveled on the second gallon, and so on, with Y20 being the number of miles traveled on the 20th gallon. Let X be the average content in a random sample of 10 rocks. The joint probability mass function is not equal to the product of the marginals.

Let A be the event that component A functions, let B be the event that component B functions, let C be the event that component C functions, and let D be the event that component D functions. Let A denote the event that the resistance is above specification, and let B denote the event that the resistance is below specification. Then A and B are mutually exclusive. Let R be the event that the shipment is returned. Let B1 be the event that the first brick chosen meets the specification, let B2 be the event that the second brick chosen meets the specification, let B3 be the event that the third brick chosen meets the specification, and let B4 be the event that the fourth brick chosen meets the specification.

Since the sample size of 4 is a small proportion of the population, it is reasonable to treat these events as independent, each with probability 0. Let A be the event that the bit is reversed at the first relay, and let B be the event that the bit is reversed at the second relay. Let A be the event that two different numbers come up, and let B be the event that one of the dice comes up 6.

Then A contains 30 equally likely outcomes 6 ways to choose the number for the first die times 5 ways to choose the number for the second die.

Of these 30 outcomes, 10 belong to B, specifically 1,6 , 2,6 , 3,6 , 4,6 , 5,6 , 6,1 , 6,2 , 6,3 , 6,4 , and 6,5. Let D denote the event that a snowboard is defective, let E denote the event that a snowboard is made in the eastern United States, let W denote the event that a snowboard is made in the western United States, and let C denote the event that a snowboard is made in Canada.

Each is equally likely to be chosen. Of these pairs, five are adjacent 1 and 2, 2 and 3, 3 and 4, 4 and 5, 5 and 6. The marginal probability mass function pX x is found by summing along the rows of the joint probability mass function. For tensile strength Y : The marginal probability mass function pY y is found by summing down the columns of the joint probability mass function. It is reasonable to assume that knowledge of the outcome of one coin will not help predict the outcome of the other.

There are nine such ordered pairs, and each is equally likely. This occurs if none of the individuals has the disease. We can get a more accurate estimate by subtracting the bias of Let n be the necessary number of measurements.

Let X represent the estimated mean annual level of land uplift for the years —, and let Y represent the estimated mean annual level of land uplift for the years — The volume is estimated with X.

Section 3. Reducing the uncertainty in P1 to 0. Reducing the uncertainty in p to 0. Reducing the uncertainty in L to 0. Reducing the uncertainty in k to 0. It is probably not worthwhile to implement the new procedure for a reduction this small. Reducing the uncertainty in b to 0. R is the only variable that substantially affects the uncertainty in Y. R k l d The relative uncertainty is 5.

Supplementary Exercises for Chapter 3 1. Reducing the uncertainty in H to 0. Reducing the uncertainty in the mass to 0. Let X1 and X2 represent the estimated thicknesses of the outer layers, and let Y1 ,Y2 ,Y3 ,Y4 represent the estimated thicknesses of the inner layers.

The thickness is 5. Reducing the uncertainty in l to 0. Therefore the estimated strength is 80, pounds in both cases. Let s denote the length of a side of the square. Since S and C are both computed in terms of s, they are not independent. The difference between the two estimates is much less than the uncertainty.

A Bernoulli random variable has possible values 0 and 1. The possible values of Y are 0 and 2. Therefore Z is a Bernoulli random variable. Section 4. Let X be the number of automobiles that violate the standard. Let X be the number of failures that occur in the base metal.

Let X be the number of women among the five winners. Let X be the number of defective parts in the sample from vendor A, and let Y be the number of defective parts in the sample from vendor B.

Let pA be the probability that a part from vendor A is defective, and let pB be the probability that a part from vendor B is defective. The observed value of X is 12 and the observed value of Y is Consulting Table A. Let p be the probability that a tire has no flaw. The results of Example 4. Let X be the number of tires out of four that have no flaw. Let q be the probability that exactly one of four tires has a flaw. Let X be the number of messages that fail to reach the base station. Now let Y have a Poisson distribution with mean 3.

The variance of Y is also equal to 3, because the variance of a Poisson random variable is always equal to its mean. Therefore Y has a larger variance than X. Let X represent the number of bacteria observed in 0. Since the mean number of flaws is 0.

If the mean concentration is 7 particles per mL, then only about 7 in every thousand 1 mL samples will contain 1 or fewer particles. Let X be the number of flaws in a one-square-meter sheet of aluminum. Let p be the probability that a one-square-meter sheet of aluminum has exactly one flaw. From Example 4. Let X be the number of units with broken fans among the seven chosen. Let X be the number of the day on which the computer crashes.

Let X be the number of hours that have elapsed when the fault is detected. Let X1 denote the number of orders for a 2. Alternatively, use Table A. Therefore a lifetime of Therefore a score of is on the 91st percentile, approximately.

Therefore a score of is on the 89th percentile, approximately. The process will be shut down on 3. Since a process with this broth will be shut down on 2. Let X be the diameter of a hole, and let Y be the diameter of a piston. The z-score of 0. To find the probability that the clearance will be between 0.

Therefore P 0. The proportion is 0. This value must be chosen so that the 1st percentile of the distribution is Therefore 65 is the 10th percentile of strength. Equation 4. Let X be the lifetime of bulb A and let Y be the lifetime of bulb B. Since D is a linear combination of independent normal random variables, D is normally distributed.

Since S is a linear combination of independent normal random variables, S is normally distributed. Therefore 0. Therefore only 0. In fact, the weakest bolt in part c is stronger than the weakest bolt in part a , the second-weakest bolt in part c is stronger than the second-weakest bolt in part a , and so on. This sample contains an outlier , so the normal distribution should not be used. Let Y be the lifetime of the component. The z-score of 1. Therefore the z-score of ln y90 must be 1.

Let Y represent the BMI for a randomly chosen man aged 25— The z-score of 3. Therefore the z-score of ln y75 must be 0. Let X represent the withdrawal strength for a randomly chosen annularly threaded nail, and let Y represent the withdrawal strength for a randomly chosen helically threaded nail.

Therefore the probability for annularly threaded nails is 0. Now find the probability for helically threaded nails. Annularly threaded nails have the greater probability. The probability is 0. Let m be the median of X. For annularly threaded nails, the z-score of 2. Therefore a strength of 20 is extremely small for an annularly threaded nail; less than one in ten thousand such nails have strengths this low.

For helically threaded nails, the z-score of 2. Therefore about 4. We can be pretty sure that it was a helically threaded nail. Only about 0. Let X represent the price of a share of company A one year from now. Let Y represent the price of a share of company B one year from now. It follows that P is lognormal. Let X be the diameter in microns. The probability that a diameter is less than 3 microns is 0. The probability that a diameter is greater than 11 microns is 0. If the lifetimes were exponentially distributed, the proportion of used components lasting longer than 5 years would be the same as the proportion of new components lasting longer than 5 years, because of the lack of memory property.

Let T be the waiting time between accidents. Let T be the waiting time. Let X be the number of mornings on which the waiting time is less than 5 minutes.

Let T be the lifetime in hours of a bearing. Let T be the lifetime of a fan. By definition, an estimator is unbiased if its mean is equal to the true value. The MLE is the value of p that maximizes f x; p , or equivalently, ln f x; p. The MLE of p is the value of p that maximizes f x; p , or equivalently, ln f x; p. The joint probability density function of X1 , If the logs of the PM data come from a normal population, then the PM data come from a lognormal population, and vice versa.

The z-score of Therefore x80 satisfies the equation 0. From the results of Example 4. Let X be the number of wires in a sample of that have no flaws. Let n be the required number of measurements. Let X be the average of the n measurements. From part a the probability that a shipment is acceptable is 0. Using the continuity correction, This equation can be rewritten as , Now X is approximately normal with mean and variance , and Y is approximately normal with mean and variance Now X is approximately normal with mean 40 and variance 40, and Y is approximately normal with mean 50 and variance Let X be the number of particles withdrawn in a volume of v mL.

The average number of particles per mL will be between 48 and 52 if the total number of particles X is between 48v and 52v. Only about 2 in 10, samples of size will have 75 or more nonconforming tiles if the goal has been reached. Let X be the number of rivets from vendor A that meet the specification, and let Y be the number of rivets from vendor B that meet the specification.

Let X be the number of people out of who appear for the flight. Let X denote the number of devices that fail. This area is 0. The 90th percentile of a normal distribution is 1. This equation can be rewritten as 62, Almost none of the samples will have a total weight of The level is the proportion of samples for which the confidence interval will cover the true value. Therefore as the level goes up, the reliability goes up. This increase in reliability is obtained by increasing the width of the confidence interval.

Therefore as the level goes up the precision goes down. The level is 0. The sample size needs to be The upper confidence bound 0. The confidence interval is for the population mean, not the sample mean. The sample mean is known, so there is no need to construct a confidence interval for it. The standard deviation of the mean involves the square root of the sample size, not of the population size.

The supervisor is underestimating the confidence. Section 5. Let p be the population proportion of components that are defective. Let r represent the proportion of lots that are returned. Extra precision is used for this confi- dence interval to get good precision in the final answer.

Substituting 0. The confidence interval is 0. Since the upper limit is greater than 1, replace it with 1. Then n satisfies the equation 0. Yes it is appropriate, since there are no outliers. Alternatively, one may compute 2. The minimum possible value is 0, which is less than two sample standard deviations below the sample mean. Therefore it is impossible to observe a value that is two or more sample standard deviations below the sample mean.

This suggests that the sample may not come from a normal population. Since 0 is in the confidence interval, it may be regarded as being a plausible value for the mean difference in hardness. It is not possible. The amounts of time spent in bed and spent asleep in bed are not independent. If additional patients are p treated with bare metal stents, the standard deviation of the difference be- tween the proportions is then 0.

If additional patients areptreated with drug-coated stents, the standard deviation of the difference be- tween the proportions is then 0.

If additional patients are treated with bare metal stents and additional patients are treated with drug- coated p stents, the standard deviation of the difference between the proportions is then 0. Therefore the confidence interval would be most precise if new patients are treated with drug-coated stents. The sample proportions come from the same sample rather than from two independent samples.

No, these are not simple random samples. The differences are: 9, 7, 5, 9, 10, 9, 2, 8, 8, The differences are: 25, 32, 39, 23, The differences are: 8. Since the values 0. Supplementary Exercises for Chapter 5 1. The higher the level, the wider the confidence interval. Therefore the narrowest interval, 4. Let n be the required sample size. This a specific confidence interval that has already been computed. The notion of probability does not apply. The confidence interval specifies the location of the population mean.

It does not specify the location of a sample mean. It does not specify the location of a future measurement. The estimate of V is approximately normal. Since 0. Therefore we are convinced that the mean daily output is not tons or more, but is instead less than tons.

The null distribution specifies that the population mean, which is also the mean of X, is the value on the boundary between the null and alternate hypotheses. Section 6. The larger the P-value, the more plausible the null hypothesis. A P-value of 0. This is unlikely, but not impossible. Since the decrease is statistically significant, it is reasonable to conclude that the homicide rate did actually decrease.

However, we cannot determine what caused the decrease. No, she cannot conclude that the null hypothesis is true, only that it is plausible. For either of the other two hypotheses, the P-value would be 0. The value 3. Quantities greater than the upper confidence bound will have P-values less than 0. The z-score is 3. The P-value is 0. The claim is rejected.

We cannot conclude that more than half of bathroom scales underestimate weight. We can conclude that more than half of residents are opposed to building a new shopping mall. We cannot conclude that the locus is not in Hardy-Weinberg equilibrium. From the t table, 0. We conclude that the scale may well be calibrated correctly. We cannot conclude that the mean surface deflection is greater than 3 mm. We can conclude that the mean surface deflection is greater than 2.

It cannot be concluded that the mean thickness differs from 4 mils. We can conclude that the mean hospital stay is shorter for patients receiving C4A-rich plasma. We can conclude that the mean dielectric constant differs between the two types of asphalt.

We cannot conclude that the mean reduction from drug B is greater than the mean reduction from drug A. We cannot conclude that the mean score on one-tailed questions is greater. We cannot conclude that the mean score on one-tailed questions differs from the mean score on two-tailed questions.

We can conclude that the mean weight loss is greater for those on the low-carbohydrate diet. We cannot conclude that the mean weight loss on the low-carbohydrate diet is more than 1 kg greater than that of the low-fat diet. We cannot conclude that the mean number of hours per week increased between and Therefore machine 1 should be used. We can conclude that the response rates differ between public and private firms. We can conclude that the proportion of patch users is greater among European-Americans.

The evidence suggests that heavy packaging reduces the proportion of damaged shipments, but may not be conclusive. We cannot conclude that the proportion of wells that meet the standards differs between the two areas.

We cannot conclude that the proportions differ between the two groups. No, because the two samples are not independent. We can conclude that the mean strength of crayons made with dye B is greater than that made with dye A. We cannot conclude that the mean strength of crayons made with dye B exceeds that of those made with dye A by more than 1 J. We can conclude that the resilient modulus for rutted pavement differs from that of nonrutted pavement. We cannot conclude that the calibration has changed from the first to the second day.

We cannot conclude that the mean cost of the new method is less than that of the old method. We cannot conclude that the mean compressive stress is greater for no. We can conclude that the mean load at failure is greater for rounded connections than for double-rounded connections. We can conclude that the mean time to sleep is less for the new drug. We can conclude that there is a difference in latency between motor point and nerve stimulation. We can conclude that the etch rates differ between center and edge.

We can conclude that the mean amount of corrosion differs between the two formulations. The differences are 35, 57, 14, 29, We can conclude that the mean strength after 6 days is greater than the mean strength after 3 days. We cannot conclude that the mean response differs between the two drugs. We can conclude that the mean number of miles per gallon is higher with radial tires. We can conclude that the mean mileage with radial tires is more than 2 miles per gallon higher than with bias tires.

We cannot conclude that the mean conversion is less than We can conclude that the mean conversion is greater than We can conclude that the mean conversion differs from We cannot conclude that the gauges differ.

The test statistic W is the sum of the ranks corresponding to the X sample. We can conclude that the mean recovery times differ. Since n and m are both greater than 8, compute the z-score of W and use the z table. We cannot conclude that the mean scores differ. The expected values are np1 , np2 and np3 , or , 50, and The row totals are O.

The column totals are O1. The grand total is O.. We can conclude that the proportions in the various income categories differ between men and women. The row totals are O1. The column totals are O. There is some evidence that the proportions of workers in the various disease categories differ among exposure levels. There is no evidence that the rows and columns are not independent. Let p1 represent the probability that a randomly chosen plate is classifed as premium, let p2 represent the probability that it is conforming, let p3 represent the probability that it is downgraded, and let p4 represent the probability that it is unacceptable.

The expected values are np1 , np2 , np3 , and np4 , or 20, , 30, and The observed values are 19, , 35, and We can conclude that the proportion of false alarms whose cause is known differs from month to month. Therefore the P-value is 0. The numbers of degrees of freedom are 8 and Since this is a two-tailed test, the P-value is twice the area to the right of 1. We cannot conclude that the variance of the breaking strength varies between the composites.

H0 is rejected at any level greater than or equal to 0. The rejection region should consist of values for X that will make the P-value of the test less than or equal to a chosen threshold level. Under H0 , the z-score of The rejection region should consist of values for X that will make the P-value of the test less than a chosen threshold level. This rejection region contains values of X greater than Therefore H0 is true, so rejecting it is a type I error.

H0 is false and was rejected. H0 is true and was not rejected. H0 is false and was not rejected. The error in question is rejecting H0 when true, which is a type I error.

When the level is smaller, it is less likely to reject H0 , and thus less likely to make a type I error. This is the definition of power.

When H0 is false, making a correct decision means rejecting H0. The power is 0. The power is not the probability that H0 is true. If the level increases, the probability of rejecting H0 increases, so in particular, the probability of rejecting H0 when it is false increases.

Therefore the level is 0. Therefore the power is 0. The z-score of 49, To learn more, view our Privacy Policy. To browse Academia. Log in with Facebook Log in with Google. Remember me on this computer. Enter the email address you signed up with and we'll email you a reset link. Need an account? Click here to sign up. Download Free PDF. Imtiaz Mustafa. A short summary of this paper. The student solution manual provides worked solutions and answers to only the odd-numbered problems given at the end of the chapter sections.

In addition to the material contained in the student solution manual, this instructor manual therefore provides worked solutions and answers to the even-numbered problems given at the end of the chapter sections together with all of the supplementary problems at the end of each chapter.

This probability is less than 0. No, because the ace of hearts could be drawn. Event B is a necessary condition for event A and so conditioning on event B increases the probability of event A. Consequently, the probability that a birthday falls on March 1st. Consequently, the probability that a birthday falls on February 1st. Similar to part b. To show that two events are independent it needs to be shown that one of the above three conditions holds.

The branches are independent since the switches operate independently of each other. This probability is equal to 0. The smallest values of n for which the probability is greater than 0. Note that in these calculations it has been assumed that birthdays are equally likely to occur on any day of the year, although in practice seasonal variations may be observed in the number of births.

Then the result 1,2,3,4 indicates that each competitor received the same rank- ing in both tournaments. The number of ways in which the best 5 competitors can be chosen is 20!

Ckn is the number of ways that k balls can be selected from n balls. Either all k balls selected are blue or one of the selected balls is red. If the line is made into a circle and rotations of the circle are considered to be indistinguishable, then there are n arrangements of the line corresponding to each arrangement of the circle.

Consequently, there are n! See the previous problem. The number of ways that six people can sit around a dinner table is 5! At the cinema these 5 blocks can be arranged in 5! If Andrea refuses to sit next to Scott then the number of seating arrangements can be obtained by subtraction.

Thus, there is a one-to-one correspondence between the positioning of the colored balls in part b and the arrangements of the balls in part a so that the problems are identical.

The total number of possible samples is C12 a The number of samples containing only items which have either excellent or If one obtains a head and the other a tail, then the one with the head wins this could just as well be done the other way around so that the one with the tail wins, as long as it is decided beforehand.

If both captains obtain the same result, that is if there are two heads or two tails, then the procedure could be repeated until different results are obtained. There are 36 equally likely outcomes, 16 of which have scores differing by no more than one.

The number of ways to pick a diamond picture card is 3. Let the notation x, y indicate that the score on the red die is x and that the score on the blue die is y. The total number of possible samples is C10 If the first and the second job are assigned to production line II, the number of assignments is 18!

If the first and the second job are assigned to production line III, the number of assignments is 18! Without replacement: xi 0 1 2 pi 0. It could be considered to be either discrete or continuous. If you can play the game a large number of times then you should play the game as often as you can. Increasing the probability of scoring a three reduces the expected value of the difference in the scores of the two dice.

R 50 A small variance is generally preferable if the expected winnings are positive. If they were independent it would suggest that one of the inspectors is randomly assigning a safety score without paying any attention to the actual state of the building.

The closer the correlation is to one the more consistent the inspectors are. R2 R2 R2 3xyz 2 2. However, if the drawings are made without replacement then the random variables Xi are not independent. This problem illustrates that the variability of the return on an investment can be reduced by diversifying the investment, so that it is spread over several funds.

The interquartile range is 9. The profits may or may not be independent depending on the type of insurance and the pool of customers.

If large natural disasters affect the pool of customers all at once then the claims would not be independent.

One head and one tail gives a total score of Two tails gives a total score of Each of these r components has a geometric distribution with parameter p. If the first two cards are spades then the probability that the first heart card is obtained on the fifth drawing is the same as the probability in part a.

Therefore, P B , 0. The probability that an order is received over the internet and it is small is 0. The probability that an order is not received over the internet and it is small is 0. The answer is 2! The probability that team A wins the series is 0. The required probability is P B 25, 0. The probability that the total waiting time is longer than 15 minutes is the probability that the additional waiting time is longer than 10 minutes, which is 0.

However, in this case the additional waiting time has a U 0, 15 distribution. Using the negative binomial distribution, the probability that exactly ten cultures are opened is! Using the multinomial distribution, the required answer is 10! Chapter 5 The Normal Distribution 5. The interquartile range of a N 0, 1 distribution is therefore 0. Solving P N 4. Solving P N P N Since z0.

The lower quartile is e5. Therefore, the airline can book up to passengers on the flight. The required probability is! Consequently, operator B started working at am. You have to consider whether the eye colors of computing students are representative of the eye colors of all students or of all people.

Perhaps eye colors are affected by race and the racial make-up of the class may not reflect that of the student body or the general public as a whole. The conditions depend upon how representative the period between and on that Saturday afternoon is of other serving periods. The service times would be expected to depend on how busy the restaurant is and on the number of servers available.

The spacing of the sampling times should ensure that the sample is representative. The random selection of the sample should ensure that it is representative of that particular delivery of bricks. However, that specific delivery of bricks may not be representative of all of the deliveries from that company.

The selection method of the sample should ensure that it is representative. If the 80 sample panels are selected in some random manner then they should be representative of the entire population.

In addition, the second largest observation 51 might also be considered to be an outlier. A test of the fairness of the die could be made using the methods presented in section The observations , , , and can all be considered to be outliers. The smallest observations 6. To avoid confusion, the answers given here assume that no observations have been removed. The sample median is The sample trimmed mean is The upper sample quartile is The lower sample quartile is The sample median is 1.

The sample trimmed mean is 1. The upper sample quartile is 1. The lower sample quartile is 1. The sample median is 3. The sample trimmed mean is 3. The upper sample quartile is 5.

The lower sample quartile is 2. The sample median is 2. The sample trimmed mean is 2. The upper sample quartile is 4. The sample median is 0. The sample trimmed mean is 0. The upper sample quartile is 0. The lower sample quartile is 0. The sample median is 9. The sample trimmed mean is 9. The upper sample quartile is 9. The lower sample quartile is 8. However, the sample may not be representative if some species are more likely to be observed than others. It appears that the grey markings are the most common, followed by the black markings.

The sample is skewed. The most frequent data value the sample mode is one error. The upper sample quartile is 2. This could be representative of all adult males in the population unless there is something special about the clientele of this clinic. The largest observation If the soil is of a fairly uniform type, and if the samples were taken representa- tively throughout the site, then they should provide accurate information on the soil throughout the entire construction site.

The likelihood is L x1 ,. This probability is 0. The confidence interval does not contain the value 2. The confidence interval contains the value 1.

A sample size slightly larger than 20 should be sufficient. The confidence interval does not contain the value The confidence interval contains the value 0.

The confidence interval contains the value Since 1. The value 3. Note: The sample statistics for the following problems in this section and the related problems in this chapter depend upon whether any observations have been removed as outliers. There is sufficient evidence to conclude that the machine is miscalibrated. There is sufficient evidence to conclude that the chemical plant is in violation of the working code. There is not sufficient evidence to conclude that the advertised claim is false.

There is sufficient evidence to conclude that the average corrosion rate of chilled cast iron of this type is larger than 2. There is some evidence that the manufacturing process needs adjusting but it is not overwhelming. There is not sufficient evidence to conclude that the spray painting machine is not performing properly.

There is sufficient evidence to conclude that the design criterion has not been met. There is not sufficient evidence to conclude that the average voltage of the batteries from the production line is at least There is sufficient evidence to conclude that the average length of the components is not There is some evidence to conclude that the components have an average weight larger than 70, but the evidence is not overwhelming.

There is sufficient evidence to conclude that the average breaking strength is not 7. There is sufficient evidence to conclude that average failure time of this kind of component is at least 50 hours. Consequently, the experiment does not provide sufficient evidence to conclude that the average time to toxicity of salmon fillets under these storage conditions is more than 11 days. There is some evidence that the average distance at which the target is detected is at least 65 miles although the evidence is not overwhelming.

The company can safely conclude that the telephone surveys will last on average less than ten minutes each. There is not sufficient evidence to conclude that the paper does not have an average weight of There is not sufficient evidence to establish that the average deformity value of diseased arteries is less than 0. Since 2. There is sufficient evidence to conclude that the average weight gain for com- posites of this kind is smaller than 0.

There is sufficient evidence to conclude that the average soil compressibility is no larger than There is sufficient evidence to conclude that the average clay compressibility at the location is less than There is not sufficient evidence to conclude that the average strength of fibers of this type is at least The critical points in Table III imply that the p-value is larger than 0.

There is some evidence that the average resistance of wires of this type is not There is not sufficient evidence to establish that the population average is not There is sufficient evidence to conclude that the flexibility of this kind of metal alloy is smaller than P9 xi This large p-value indicates that H0 should be accepted. This small p-value indicates that H0 should be rejected. There is some evidence to establish that the average density of these kind of compounds is larger than 3.

There is sufficient evidence to establish that the population mean is not The critical points in Table III imply that the p-value is between 0. The critical points in Table III imply that the p-value is smaller than 0. Chapter 9 Comparing Two Population Means 9. There is not sufficient evidence to conclude that the new assembly method is any quicker on average than the standard assembly method.

With t0. There is sufficient evidence to conclude that the new tires are better than the standard tires in terms of the average reduction in tread depth. There is sufficient evidence to conclude that the new teaching method is better since it produces higher scores on average than the standard teaching method. There is not sufficient evidence to conclude that the two laboratories are any different in the datings that they provide. There is not sufficient evidence to conclude that the new golf balls travel further on average than the standard golf balls.

There is not sufficient evidence to conclude that procedures A and B give different readings on average. Consequently, there is some evidence that the new antibiotic is quicker than the standard antibiotic, but the evidence is not overwhelming. Consequently, there is not sufficient evidence to conclude that the addition of the surfactant has an effect on the amount of uranium-oxide removed from the water.