ANIMSCI 2260 Exam II

You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 years of ownership. Since you are particulary interested in a certain foreign sedan, you decide to estimate the resale value of this car with a 95% confidence interval. You manage to obtain data on 16 recently resold 5-year-old foreign sedans of that model. These 16 cars were resold at an average price of $10,000 with a standard deviation of $1,000. Estimate the true mean resale value of this model of foreign car using a 95% confidence interval. ($8,000, $12,000) ($9,470, $1,530) ($9,510, $10,490) ($9,467.25, $10,532.75)

($9,467.25, $10,532.75)

Psychologists at the University of Minnesota compared the scores of independent random samples of male and female 8th grade students who took a basic skills math achievement test to determine if males outperform females in math. A summary of the test score data is shown below: Males Females Sample size 1,764 1,739 Sample means 48.9 48.4 Sample standard deviations 12.96 11.85 Use a 90% confidence interval to estimate the true difference in mean test scores between males and females. (-0.19004, 1.19004) (-0.32218, 1.32218) (24.30996, 25.69004) (0.41996, 1.80004)

(-0.19004, 1.19004)

A study published in The Journal of American Academy of Business examined whether the perception of service quality at five-star hotels in Jamaica differs by gender. In order to compare the means of two populations (i.e., male vs. female guests), independent random samples were selected from each population, with the results shown in the table below. Use these data to construct a 96% confidence interval for the difference in the two population means. Males Females Sample size 130 115 Sample mean score 39.10 38.70 Sample standard deviation 6.70 6.95 (-1.393404, 2.193404) (0.05667, 0.75243) (-0.255855, 1.055855) (-0.285971, 1.085971)

(-1.393404, 2.193404)

An instructor wants to estimate the proportion of students who have received an A in the statistics course he has taught for the past 10 yr. He selects a random sample of 100 student records and finds that 15 of the 100 students received an A for the course. Perform the calculations needed to determine whether or not it would be appropriate to construct a large-sample confidence interval.

(0.04288, 0.25712) p̂ ± 3σp̂ = p̂ ± 3 * [√(p̂q̂/n)] = 0.15 ± 3 * [√( (0.15 * 0.85)/100 )] = 0.15 ± 3 * [√0.001275] = 0.15 ± 3 * (0.0357071) = 0.15 ± 0.10712 = (0.04288, 0.25712) This interval does not include 0 or 1.0. Therefore, it is appropriate to use large-sample confidence interval procedures.

An instructor wants to estimate the proportion of students who have received an A in the statistics course he has taught for the past 10 yr. He selects a random sample of 100 student records and finds that 15 of the 100 students received an A for the course. Construct a 96% confidence interval for the true population proportion (p) of students receiving an A in this class.

(0.07644, 0.22356) = 0.15 ± z(0.04/2) * [√( (0.15 * 0.85)/100 )] = 0.15 ± 2.06 * 0.0357071 = 0.15 ± 0.0735566 = (0.07644, 0.22356)

An instructor wants to estimate the proportion of students who have received an A in the statistics course he has taught for the past 10 yr. He selects a random sample of 100 student records and finds that 15 of the 100 students received an A for the course. Construct a 90% confidence interval for the true population proportion (p) of students receiving an A in this class.

(0.09126, 0.20874) p̂ ± z(α/2) * [√(p̂q̂/n)] = 0.15 ± z(0.10/2) * [√( (0.15 * 0.85)/100 )] = 0.15 ± 1.645 * 0.0357071 = 0.15 ± 0.0587381 = (0.09126, 0.20874)

An animal scientist is interested in determining the proportion of ewes that give birth to twins. Rather than examine the records for all ewes in the United States, he randomly selects 500 ewes and finds that 220 of them gave birth to twins. Construct a 99% large-sample confidence interval to estimate the true population proportion of ewes who give birth to twins. (0.39649, 0.48351) (191.41867, 248.58133) (0.3828, 0.4972) (198.24489, 241.75511)

(0.3828, 0.4972)

A random sample of 4,000 U.S. citizens was asked their opinion concerning gun control. A total of 2,600 of the 4,000 citizens in the sample stated that they are in favor of gun control legislation. Construct a 98% confidence interval for the true population proportion (p). (0.63243, 0.66757) (0.62738, 0.67262) 0.65 (2,529. 71274, 2,670.28726)

(0.63243, 0.66757)

In order to compare the means of two populations, independent random samples are selected from each population, with the following results: Sample 1 Sample 2 Sample size 500 400 Sample mean 5,280 5,240 Sample standard deviation 150 200 Construct a 95% confidence interval for the difference in the two population means. (37.915985, 42.0840151) (16.398484, 63.601516) (11.94308, 68.05692) (-3.92777, 9.92777)

(16.398484, 63.601516)

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in Franklin County. The milk production records have a normal distribution. We select a random sample of milk production records for 25 Holstein cows. Note that this is a small sample (i.e., n < 30). The sample mean is 20,000 lb of milk and the sample standard deviation is 2,000 lb of milk. Construct a 99% confidence interval for the population mean (μ).

(18881.2, 21118.8) = 20000 ± 1118.8 = (18881.2, 21118.8)

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in Franklin County. The milk production records have a normal distribution. We select a random sample of milk production records for 25 Holstein cows. Note that this is a small sample (i.e., n < 30). The sample mean is 20,000 lb of milk and the sample standard deviation is 2,000 lb of milk. Construct a 95% confidence interval for the population mean (μ).

(19174.4, 20825.6) = 20000 ± 2.064 * (400) = 20000 ± 825.6 = (19174.4, 20825.6)

We want to estimate the mean yield of soybeans in Ohio in the past year. Therefore, we obtain the yields of a random sample of 100 soybean fields. The average yield of the samples is 40 bushels per acre and the standard deviation of the yields is 5 bushels per acre. Construct a 95% confidence interval for the population mean (μ).

(39.02, 40.98) = 40 ± z(0.05/2) * (5 / √100) = 40 ± 1.96 * (0.5) = 40 ± 0.98 = (39.02, 40.98)

We want to estimate the mean yield of soybeans in Ohio in the past year. Therefore, we obtain the yields of a random sample of 100 soybean fields. The average yield of the samples is 40 bushels per acre and the standard deviation of the yields is 5 bushels per acre. Construct a 90% confidence interval for the population mean (μ).

(39.1775, 40.8225) x̄ ± z(α/2) * (s / √n) = 40 ± z(0.10/2) * (5 / √100) = 40 ± 1.645 * (0.5) = 40 ± 0.8225 = (39.1775, 40.8225)

Suppose we want to estimate the average height of all students enrolled in Animal Sciences 2260 this semester. We select a random sample of 25 students and measure their heights. The sample mean is 68 inches and the sample standard deviation is 10 inches. Which one of the following is the correct 95% confidence interval for the population mean? (64.08 inches, 71.92 inches) (47.36 inches, 88.64 inches) (64.578 inches, 71.422 inches) (63.872 inches, 72.128 inches)

(63.872 inches, 72.128 inches)

Independent random samples, each containing 800 observations, were selected from two binomial populations. The samples from populations 1 and 2 produced 320 and 400 successes, respectively. Find the point estimate of the true difference in population proportions of successes. -0.10 0.90 0.45 -80

-0.10

Find the probability that a normally distributed random variable will lie more than z = 3.0 standard deviations above its mean.

0.0013 P (z > 3.0) = 0.50 - 0.4987 = 0.0013

Suppose x has a binomial probability distribution with n = 50 and p = 0.60. Use the normal approximation to the binomial to find P (X≤20).

0.0031 P (X≤20) ≈ P [ z < { ( (20+0.5)-(50 * 0.60) ) / ( √(50 * 0.60 * 0.40) ) } ] ≈ P [ z < { (20.5 - 30) / 3.4641016 } ] ≈ P [ z < -2.74 ] P (z < -2.74) = 0.50 - 0.4969 = 0.0031

Suppose x has a binomial probability distribution with n = 50 and p = 0.60. Use the normal approximation to the binomial to find P (X≥40).

0.0031 P (X≥40) ≈ P [ z > { ( (40 - 0.5)-(50 * 0.60) ) / ( √(50 * 0.60 * 0.40) ) } ] ≈ P [ z > { (39.5 - 30) / 3.4641016 } ] ≈ P [ z > 2.74 ] P (z > 2.74) = 0.50 - 0.4969 = 0.0031

The mean weight in a herd of pigs is 220 lb and the standard deviation of the weights is 40 lb. What proportion of the pigs would be expected to weigh more than 300 lb?

0.0228 z = (300-220)/40 = 2.0 P (x > 300) = P (z > 2.0) = 0.50 - 0.4772 = 0.0228

Find the probability of an observation lying more than z = 1.77 standard deviations above the mean. 0.4616 0.9616 0.0384 2.0

0.0384

The average height of a certain ornamental plant is 14 inches and the standard deviation of the heights is 2 inches. Find the probability that a randomly selected plant will have a height of more than 17.5 inches. 0.1250 1.75 0.0401 0.4599

0.0401

A physical fitness association is including the mile run in its secondary-school fitness test for boys. The time for this event for boys in secondary school is known to have a normal distribution with a mean of 450 seconds and a standard deviation of 50 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 370 seconds. 0.9452 0.0548 0.4452 1.60

0.0548

Find the probability of an observation lying more than z = 1.56 standard deviations below the mean. 0.4406 0.0594 0.9406 0.0256

0.0594

The mean weight in a herd of pigs is 220 lb and the standard deviation of the weights is 40 lb. What proportion of the pigs would be expected to weigh less than 160 lb?

0.0668 z = (160-220)/40 = -1.5 P (x < 160) = P (z < -1.5) = 0.50 - 0.4332 = 0.0668

Find the area under the normal curve between z = 1.3 and z = 1.9

0.0681 P (1.3 < z < 1.9) = 0.4713 - 0.4032 = 0.0681

Suppose x has a binomial probability distribution with n = 200 and p = 0.60. Use the normal approximation to the binomial to find P (X > 130). 1.37 0.0853 0.4147 0.2578

0.0853

Scientists want to estimate the difference in twinning rate of two lines of beef cattle that have been selected for increased frequency of twin births. Last spring, 40 of the 100 cows in Line 1 gave birth to twins. In Line 2, 30 of the 100 cows gave birth to twins. Find the point estimate of the true difference in population proportions of cows giving birth to twins in Lines 1 and 2. 0.8333 0.35 0.10 10

0.10

The average height of a certain ornamental plant is 15 inches and the standard deviation of the heights is 3 inches. Find the probability that a randomly selected plant will have a height of more than 18.75 inches. 1.25 0.1056 0.8944 0.3255

0.1056

An instructor wants to estimate the proportion of students who have received an A in the statistics course he has taught for the past 10 yr. He selects a random sample of 100 student records and finds that 15 of the 100 students received an A for the course. What is the point estimate of the true population proportion of students receiving an A in this class?

0.15 p̂ = point estimate = 15/100 = 0.15

An animal scientist conducts a study to estimate the proportion of cows of the Charolais breed that require assistance during calving. A random sample of 500 Charolais cows is selected. Results of the study show that 100 of the 500 cows in the sample required assistance in giving birth. What is the point estimate of the true population proportion (p) of Charolais cows that require assistance during calving? 100 0.20 0.10 0.25

0.20

Find the area under the normal curve between the mean and a point z = -1.47 to the left of the mean

0.4292

The mean weight in a herd of pigs is 220 lb and the standard deviation of the weights is 40 lb. What proportion of the pigs would be expected to weigh between 200 and 250 lb?

0.4649 z1 = (200-220)/40 = -0.50 z2 = (250-220)/40 = 0.75 P (200 < x < 250) = P (-0.50 < z <0 .75) = 0.1915 + 0.2734 = 0.4649

Find the area under the normal curve between the mean and a point z = 2.35 standard deviation units to the right of the mean.

0.4906

A random sample of 4,000 U.S. citizens was asked their opinion concerning gun control. A total of 2,600 of the 4,000 citizens in the sample stated that they are in favor of gun control legislation. What is the point estimate of the true population proportion of U.S. citizens who favor gun control legislation? 0.60 2,600 0.65 1.54

0.65

Find the probability that a normally distributed random variable will lie within z = 1.0 standard deviation of its mean.

0.6826 0.3413 - 0.3413 = 0.6826

The average height of a herd of cows is 50 inches and the standard deviation of the heights is 5 inches. Find the probability that a randomly selected cow will have a height between 44 and 58 inches. 0.40 0.8301 0.0603 0.75673

0.8301

Suppose that X has a binomial probability distribution with n = 15 and p = 0.70. Use the normal approximation to the binomial to calculate P(X>9). -1.12687 0.8708 0.1292 0.3708

0.8708

Suppose x has a binomial probability distribution with n = 50 and p = 0.60. Use the normal approximation to the binomial to find P (25≤X≤35).

0.8882 P (25≤X≤35) ≈ P [ { ( (25 - 0.5)-(50 * 0.60) ) / ( √(50 * 0.60 * 0.40) ) } < z < { ( (35 + 0.5)-(50 * 0.60) ) / ( √(50 * 0.60 * 0.40) ) } ] ≈ P [ { (24.5 - 30) / 3.4641016 } < z < { (35.5 - 30) / 3.4641016 } ] ≈ P [ -1.59 < z < 1.59 ] = 0.4441 + 0.4441 = 0.8882

The amount of corn chips dispensed into a 10 ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of 0.2 ounces. What portion of the 10 ounce bags would be expected to contain more than the advertised 10 ounces of corn chips? 0.4938 -2.50 0.0062 0.9938

0.9938

A population of rabbits has a mean weight of 12 lb with a standard deviation of 3 lb. A rabbit breeder selects 1,000 samples of 36 rabbits each from this population, calculates the mean weight of the rabbits in each of these 1,000 samples, and then graphs the 1,000 sample means. The mean of these 1,000 sample means is expected to be equal to: 0.20 lb 0.50 lb 12 lb 3.0 lb

12 lb

A random sample of 16 Standardbred horses was selected from a population of Standardbreds. The mean time required for these 16 horses to run a mile while pulling a sulky was 130 seconds (i.e., 2 minutes and 10 seconds), The standard deviation of the sample was 10 seconds. What is the point estimate of the true population mean for racing time of Standardbred horses? 10 seconds 16 0.07691 130 seconds

130 seconds

A population of cats has a mean weight of 15 lb and a standard deviation of the weights equal to 4 lb. A cat breeder selects a large number of samples of 64 cats each, calculates the mean weight of the cats in each of these samples, and then graphs the sample means. The mean of these sample means is expected to be equal to _______. 1.875 lb 15 lb 16 lb^2 0.0625 lb

15 lb

The mean weight in a herd of pigs is 220 lb and the standard deviation of the weights is 40 lb. Only 15% of the pigs would be expected to weigh less than x. Find the value of x.

178.4 lb z = (x-μ)/σ = (x-220)/40 = -1.04 X = (-1.04 * 40) + 220 = -41.6 + 220 = 178.4 lb

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in the United States. The milk production records have a normal distribution. The population mean for the milk production values of the Holstein cows is 20,000 lb of milk and the population standard deviation is 2,000 lb of milk. We select a large number of random samples of size n = 100 from this population and then plot the sample means. The mean of this distribution of sample means is expected to be equal to __________ lb.

20,000 lb μx̄ = μ = 20,000 lb

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in the United States. The milk production records have a normal distribution. The population mean for the milk production values of the Holstein cows is 20,000 lb of milk and the population standard deviation is 2,000 lb of milk. We select a large number of random samples of size n = 100 from this population and then plot the sample means. The standard deviation of this distribution of sample means is expected to be equal to __________ lb.

200 lb σx̄ = (σ / √n) = (2000 / √100) = 200 lb

The mean weight in a herd of pigs is 220 lb and the standard deviation of the weights is 40 lb. Only 30% of the pigs would be expected to weigh more than x. Find the value of x.

241 lb z = (x-μ)/σ = (x-220)/40 = 0.525 X = (0.525 * 40) + 220 = 21 + 220 = 241 lb

The mean length of time required to complete the Columbus Marathon was 4.5 hours and the standard deviation of the times was 0.50 hours. Assume that the racing times were approximately normally distributed. Only 10% of the runners would be expected to complete the race in less than x hours. Find the value of x. 3.86 hours -1.28 5.14 hours 0.40

3.86 hours

The general rule of thumb is that we need a sample size of n > _________ to use large-sample confidence interval procedures to estimate the population mean. 50 20 30 100

30

We want to estimate the mean yield of soybeans in Ohio in the past year. Therefore, we obtain the yields of a random sample of 100 soybean fields. The average yield of the samples is 40 bushels per acre and the standard deviation of the yields is 5 bushels per acre. What is the point estimate of the true population mean (μ) for the yield of soybeans in Ohio in the past year?

40 bushels/acre Point estimate of μ = x̄ = 40 bushels/acre

The average height cows of a certain breed is 54 inches and the standard deviation of the heights is 8 inches. Fifteen percent of the cows are expected to be less than X inches tall. Find the value of X. 54.4768 inches 45.68 inches -1.04 62.32 inches

45.68 inches

A crop scientist would like to know the average yield of soybeans in Ohio (in bushels per acre). A random sample of 225 soybean fields in Ohio yields a mean of 48 bushels per acre and a standard deviation of 7.5 bushels per acre. Estimate the population mean for the yield of soybeans In Ohio using a point estimate. 0.21333 (47.02 bushels/acre, 48.98 bushels/acre) 7.5 bushels/acre 48 bushels/acre

48 bushels/acre

Toyota would like to know how many miles per gallon the average person gets when driving the hybrid Toyota Prius. A random sample of 225 drivers yields a mean of 50 mpg and a standard deviation of 7.5 mpg. The point estimate of the population mean for the miles per gallon of the Toyota Prius would be: 0.5 mpg 50 mpg 0.222 mpg 4.5 mpg

50 mpg

The average height of a herd of cows is 50 inches and the standard deviation of the heights is 5 inches. Only 12% of the cows are expected to be more than X inches tall. Find the value of X. 55.875 inches 44.125 inches 51.55 inches 1.175 inches

55.875 inches

Which of the following is not one of the properties of the sampling distribution of the sample mean? The shape of the sampling distribution is approximately normal if the sample size is sufficiently large (e.g., n is greater than or equal to 30). The mean of the sampling distribution is equal to the population mean. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of n. All of the above are properties of the sampling distribution of the sample mean.

All of the above are properties of the sampling distribution of the sample mean.

An assumption required for small-sample estimation of (μ1 - μ2) is that the variances of the samples selected from the two populations are equal. True False

False

For a fixed confidence coefficient, the width of the confidence interval increases as the sample size increases. True False

False

The Central Limit Theorem guarantees that the population is normally distributed whenever n is sufficiently large (n > 30). True False

False

The standard deviation of the sampling distribution of the sample mean is equal to σ, the standard deviation of the population. True False

False

A Gallop poll is conducted to estimate the proportion of voters who plan to vote in favor of a certain issuse on the ballet. A random sample of 500 people of voting age is selected. Results of the poll show that 300 of the 500 people polled plan to vote in favor of the issue. Perform the calculations needed to determine whether or not it would be appropriate to construct a large-sample confidence interval for the true population proportion. What do you conclude (i.e., is it appropriate? It would be appropriate to construct a large-sample confidence interval for the true population proportion, because n = 500, which is greater than 30. It would be appropriate to construct a large-sample confidence interval for the true population proportion, because p-hat = 0.60, which is between 0 and 1.0. It would not be appropriate to construct a large-sample confidence interval for the true population proportion, because -1.20 and 2.4 do not fall between 0 and 1.0. It would be appropriate to construct a large-sample confidence interval for the true population proportion, because 0.5343 and 0.6657 both fall between 0 and 1.0.

It would be appropriate to construct a large-sample confidence interval for the true population proportion, because 0.5343 and 0.6657 both fall between 0 and 1.0.

Suppose x has a binomial probability distribution with n = 50 and p = 0.60. Recall that the normal approximation to the binomial will be good if both μ - 3σ and μ + 3σ fall between 0 and n. Do the calculations for this example to determine if it would be appropriate to use the normal approximation to the binomial.

It would be appropriate to use the normal approximation to the binomial. μ = np = (50 * 0.60) = 30 σ = √(npq) = √(50 * 0.60 * 0.40) = √(12) = 3.46410161514 μ - 3σ = 30 - (3 * 3.46410161514) = 30 - 10.3923048454 = 19.6076951546 μ + 3σ = 30 + (3 * 3.46410161514) = 30 + 10.3923048454 = 40.3923048454 Both 19.6076951546 and 40.3923048454 fall between 0 and n=50, so it would be appropriate to use the normal approximation to the binomial.

Assume that we have a herd of 50 horses and that we want to select a random sample of 5 of the horses for an experiment. We use a random number table to obtain the following set of random numbers: 06 11 42 42 53 Is this a legitimate random sample to use for the experiment? No. We have two number 42's in our sample and 53 is greater than 50. There is no horse number 53, because we only have 50 horses to select from. Yes

No. We have two number 42's in our sample and 53 is greater than 50. There is no horse number 53, because we only have 50 horses to select from.

Suppose that a plant scientist has available 80 plots of ground that he (she) could use for an experiment. He (she) decides to use 10 of the 80 available plots for this particular experiment. Utilize a random number table to generate a random sample of 10 plots. Explain step by step how you obtained this random sample.

Number the plots from 01, 02,...，80. Arbitrarily start at row 20, column 1 and go across the row. Ignore repeats of numbers and ignore numbers > 80. We choose the following 10 plots: 07, 33, 09, 42, 06, 76, 13, 51, 46, 19

When we use z-scores and areas under the normal curve, we are using a _______________, which has a mean of 0 and a standard deviation of 1. Uniform distribution Poisson distribution Gamma distribution Standard normal distribution

Standard normal distribution

We want to estimate the mean yield of soybeans in Ohio in the past year. Therefore, we obtain the yields of a random sample of 100 soybean fields. The average yield of the samples is 40 bushels per acre and the standard deviation of the yields is 5 bushels per acre. Which confidence interval is wider, the 90% confidence interval or the 95% confidence interval? Intuitively, does this make sense?

The 95% confidence interval is wider. It must be wider so that we can be 95% confident, rather than 90% confident, that the confidence interval contains μ.

The Central Limit Theorem says that the sampling distribution of the sample mean is approximately normal under certain conditions. Which of the following is a necessary condition for the Central Limit Theorem to be used? The population from which we are sampling must be normally distributed. The sample size must be large (at least 30 observations). The population size must be large (at least 30 observations). The population from which we are sampling must have a binomial distribution.

The sample size must be large (at least 30 observations).

The Central Limit Theorem says that the sampling distribution of the sample mean is approximately normally distributed under certain conditions. Which of the following is a necessary condition for the Central Limit Theorem to be used? The sample size must be large (i.e., n must be greater than or equal to 30). The population from which we are sampling must be normally distributed. The sample must be normally distributed. The population size must be large (i.e., n must be greater than or equal to 30).

The sample size must be large (i.e., n must be greater than or equal to 30).

Which one of the following is not an assumption required for small-sample estimation of (μ1 - μ2)? The samples selected from the two populations are normally distributed. The random samples are selected in an independent manner from the two populations. The variances of the two populations are equal. Both of the populations from which the samples are selected have relative frequency distributions that are approximately normal.

The samples selected from the two populations are normally distributed.

Which one of the following statements about the sampling distribution of the sample mean is incorrect? The sampling distribution is approximately normal whenever the sample size is sufficiently large (n> 30). The sampling distribution is generated by repeatedly taking samples of size n, computing the sample means, and then graphing these sample means. The standard deviation of the sampling distribution is equal to the population standard deviation. The mean of the sampling distribution is equal to the population mean.

The standard deviation of the sampling distribution is equal to the population standard deviation.

Which one of the following statements concerning the sampling distribution of the sample mean is incorrect? The sampling distribution is approximately normal whenever the sample size is sufficiently large (n ≥ 30). The sampling distribution is generated by repeatedly taking samples of size n, computing the sample means, and then graphing these sample means. The mean of the sampling distribution is equal to the mean of the population, μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population from which the samples were taken.

The standard deviation of the sampling distribution is equal to the standard deviation of the population from which the samples were taken.

Which one of the following is not an assumption required for large-sample estimation of (μ1 - μ2)? The two target populations have a normal distribution. The sample sizes are sufficiently large (n1 and n2 ≥ 30). The two samples are randomly selected in an independent manner from the two target populations.

The two target populations have a normal distribution.

A disadvantage of pulling numbers out of a hat as a method of random sampling is that it is difficult to get a thorough mixing of the pieces of paper in the hat or other container. Therefore, this procedure only gives an approximation of random sampling. True False

True

A disadvantage of pulling numbers out of a hat as a method of random sampling is that it is not feasible to use this method when the population consists of a large number of observations. True False

True

The t and z distributions are very similar. Both are symmetric, mound-shaped, and have a mean of zero. True False

True

The value of a sample statistic will vary from sample to sample. True False

True

Two disadvantages of using the "pulling numbers of a hat" method of obtaining a random sample are (1) that it is not feasible for large populations (e.g., 5 million cows) and (2) it is difficult to get a thorough mixing of the pieces of paper. True False

True

A large labor union wishes to estimate the mean number of hours per month that union members are absent from work. The union samples 475 of its members at random and monitors their working time for 1 month. At the end of the month, the total number of hours absent from work is recorded for each employee. The mean and standard deviation of the sample are 9.6 hours and 3.6 hours, respectively. What is the correct interpretation of a 95% confidence interval that can be used to estimate the mean (μ) of the entire population of number of hours absent from work per month? We are 95% confident that the true population proportion (p) of union workers who were absent from work will fall in the inteval that we derived. We are 95% confident that the sample mean (X-bar) falls in the interval that we derived. We are 95% confident that the true population mean (μ) falls in the interval that we derived. We expect that 95% of the values for number of hours absent from work will fall in the interval that we derived.

We are 95% confident that the true population mean (μ) falls in the interval that we derived.

It is desired to estimate the average total compensation of CEO's in the service industry. Data were collected from 18 CEO's and a 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Which one of the following interpretations is correct? 97% of the sampled total compensation values fell between $2,181,260 and $5,836,180. We are 97% confident that the average total compensation of all CEO's in the service industry falls in the interval $2,181,260 to $5,836,180. We are 97% confident that the mean of the sampled CEO's falls in the interval from $2,181,260 to $5,836,180. In the population of service industry CEO's, 97% of them will have total compensations that fall in the interval $2,181,260 to $5,836,180.

We are 97% confident that the average total compensation of all CEO's in the service industry falls in the interval $2,181,260 to $5,836,180.

A Gallop poll is conducted to estimate the proportion of voters who plan to vote in favor of a school levy in a certain school district. A random sample of 400 people of voting age is selected. Results of the poll show that 240 of the 400 people polled plan to vote in favor of the school levy. Construct a 99% confidence interval for the true population proportion of people who plan to vote in favor of the school levy. What is the meaning of this confidence interval? We are 99% confident that the true population mean falls within the interval that we derived. We are 99% confident that the true population proportion (p) falls within the interval that we derived. We are 99% confident that the sample mean falls within the interval that we derived. We are 99% confident that the sample estimate of the true population proportion falls within the interval that we derived.

We are 99% confident that the true population proportion (p) falls within the interval that we derived.

We want to estimate the proportion (p) of cows conceiving at first service in an AI program on a particular farm. How many cows would need to be included in the sample if it was desired to estimate p correct to within 0.05 with probability equal to 0.90? Assume that we have a prior estimate of p which is equal to 0.60. We need n = 198 cows We need n = 369 cows We need n = 260 cows We need 8 cows

We need n = 260 cows

Assume that a population of rabbit weights has a uniform distribution, instead of a normal distribution. We calculate the mean of 1,000 random samples from this population, where the number of observations in each sample is equal to 50. Would you expect the 1,000 sample means to be normally distributed? Yes No

Yes

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in the U.S. The milk production records have a normal distribution. We select a large number of random samples of size n = 100 from this population and then plot the sample means. Would the sample means still have a normal distribution if the population of milk production records was not normally distributed (e.g., if the population had an exponential distribution)? Yes No

Yes

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in the United States. The milk production records have a normal distribution. The population mean for the milk production values of the Holstein cows is 20,000 lb of milk and the population standard deviation is 2,000 lb of milk. We select a large number of random samples of size n = 100 from this population and then plot the sample means. Would the sample means still have a normal distribution if the population of milk production records was not normally distributed (e.g., if the population had an exponential distribution)?

Yes. According to the Central Limit Theorem, the sample means will be normally distributed regardless of the shape of the distribution from which the samples are selected.

Which one of the following confidence intervals would be the widest? a 60% confidence interval a 95% confidence interval a 98% confidence interval a 92% confidence interval

a 98% confidence interval

Suppose a 99% confidence interval for the population mean for weight of cows of a certain breed turns out to be (800 lb, 1400 lb). To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced width of the confidence interval? decrease the confidence level from 99% to 95% choices will result in a reduced width of the confidence interval decrease the sample size both of the above

decrease the confidence level from 99% to 95%

Suppose that a veterinarian has received permission from the owners of 50 dogs to use their dogs in an experiment involving a new drug for cancer treatment. The vet decides to use 5 of the 50 available dogs for a small preliminary experiment before conducting a larger study. The vet arbitrarily begins at row 8 column 1 of the random number table and goes from left to right across the row. Which one of the following is the correct random sample of 5 dogs? For your convenience, here are rows 8 and 9 of the random number table: Column Row 1 2 3 4 5 6 8 96301 91977 05463 07972 18876 20922 9 89579 14342 63661 10281 17453 18103 dogs number 96 91 05 07 18 dogs number 05 07 18 20 14 dogs number 96301 91977 05463 07972 18876 dogs number 9 6 3 0 1

dogs number 05 07 18 20 14

Suppose that a veterinarian has received permission from the owners of 60 dogs to use their dogs in an experiment involving a new drug for cancer treatment. The vet decides to use 6 of the 60 available dogs for a small preliminary experiment before conducting a larger study. The vet arbitrarily begins at row 12 column 1 of the random number table and goes from left to right across the row. Which one of the following is the correct random sample of 6 dogs? For your convenience, here are rows 12 and 13 of the random number table: Column Row 1 2 3 4 5 6 12 63553 40961 48235 03427 49626 69445 13 09429 93969 52636 92737 88974 33488 dogs number 63553 40961 48235 03427 49626 69445 dogs number 63 40 48 03 49 69 dogs number 6 3 5 5 3 4 dogs number 40 48 03 49 09 52

dogs number 40 48 03 49 09 52

Suppose that a random sample of 100 measurements is selected from a population with a mean µ = 200 lb and a variance σ2 = 1,600 lb2. What is the mean and standard deviation of the sampling distribution of the sample mean? mean = 200 lb and standard deviation = 40 lb mean = 20 lb and standard deviation = 40 lb mean = 200 lb and standard deviation = 4 lb mean = 200 lb and standard deviation = 1,600 lb2

mean = 200 lb and standard deviation = 4 lb

Suppose that a random sample of 625 measurements is selected from a population with a mean µ = 500 lb and a variance σ2 = 100 lb2. What is the mean and standard deviation of the sampling distribution of the sample mean? mean = 500 lb and standard deviation = 10 lb mean = 500 lb and standard deviation = 0.4 lb mean = 20 lb and standard deviation = 10 lb mean = 500 lb and standard deviation = 100 lb2

mean = 500 lb and standard deviation = 0.4 lb

Suppose that a random sample of 40 measurements is selected from a population with a mean of 64 and a variance of 64. What is the mean and standard deviation of the sampling distribution of the sample mean? mean = 64 and standard deviation = 1.26491 mean = 64 and standard deviation = 64 mean = 64 and standard deviation = 8 mean = 64 and standard deviation = 0.20

mean = 64 and standard deviation = 1.26491

The Central Limit Theorem states: Consider a random sample of n observations selected from any population with mean μ and standard deviation σ. If the sample size is sufficiently large, then the sampling distribution of the sample mean (X-bar) will be approximately a normal distribution with mean ________ and standard deviation _______. mean µ and standard deviation σ/√n mean µ and standard deviation σ mean of 0 and standard deviation of 1 mean X-bar and standard deviation s

mean µ and standard deviation σ/√n

We want to use a confidence interval to estimate the proportion of students in the College of Food, Agricultural, and Environmental Sciences that are female. What sample size would be necessary if we want to estimate the true proportion of female students correct to within 0.03 with probability 0.95? In an earlier small-scale pilot study, we obtained an estimate of the proportion of female students (p) that was equal to 0.48. n = 1,066 students n = 30 students n = 1,068 students n = 1,065 students

n = 1,066 students

The Lantern wants to conduct an opinion poll to estimate the true population proportion of OSU students who think OSU will win the NCAA tournament this year in basketball. How many students will the Lantern need to include in their sample to estimate the true population proportion to within 0.05 (i.e., B = 0.05) with probability equal to 0.95? No prior estimates of p and q are available. n = 384 students n = 10 students n = 271 students n = 385 students

n = 385 students

Find the sample size needed to estimate p correct to within 0.04 with probability 0.90. Assume that we have previous information available that indicates that p = 0.60. n = 224 n = 577 n = 406 n = 10

n = 406

Assuming that n1 = n2, find the sample sizes needed to estimate (p1 - p2) correct to within 0.07 with probability 0.90. Assume that there is no prior information available to obtain sample estimates of p1 and p2. n1 = n2 = 168 n1 = n2 = 12 n1 = n2 = 277 n1 = n2 = 144

n1 = n2 = 277

Assume that we are interested in the population consisting of the lactation records of all Holstein cows in the United States. The milk production records have a normal distribution. The population mean for the milk production values of the Holstein cows is 20,000 lb of milk and the population standard deviation is 2,000 lb of milk. We select a large number of random samples of size n = 100 from this population and then plot the sample means. According to the Central Limit Theorem, this plot of sample means should have a ____________ distribution.

normal

A numerical descriptive measure of a population is called a ______________. statistic parameter

parameter

A __________ __________ of a parameter is a statistic, a single value computed from the observations in a sample, that is used to estimate the value of the target parameter.

point estimate

A ______________ of n experimental units is one selected in such a way that every different sample of size n has an equal probability of being selected. random sample mean population uniform distribution

random sample

We randomly select 100,000 samples of size n from a population. We calculate the sample mean (X-bar) for each of the 100,000 random samples and graph the relative frequency distribution for these 100,000 values of X-bar. This relative frequency distribution is called the ____________________ of X-bar. mean sampling distribution theoretical distribution conditional probability distribution

sampling distribution

The American Angus Association wants to determine the proportion of their members who breed their cows using artificial insemination (AI). They randomly sample 200 of their members and ask them whether or not they breed their cows using AI. 120 of the 200 members sampled said "yes". The proportion of the 200 members who breed their cows using AI is an example of a ____________. parameter experimental unit population mean (μ) statistic

statistic

A researcher wants to determine whether men's and women's attitudes regarding environmental issues differ. Therefore, the researcher samples 100 men and 100 women and asks "Do you think the environment is a major concern"? Of those sampled, 67 women and 53 men responded that they believe that environmental issues are a major concern. What criterion is used to assess whether the Central Limit Theorem can be applied to this problem? the population proportions are equal. the interval, p-hat + 3 √(p-hat)(q-hat)/n, falls between 0 and 1 for both the men and the women. the samples are independently selected. both sample sizes are at least 30.

the interval, p-hat + 3 √(p-hat)(q-hat)/n, falls between 0 and 1 for both the men and the women.