Homework Questions (6, 7, 8)

X is a normally distributed random variable with a mean of 5 and a variance of 4. The probability that x is greater than 10.52 is (Use the Standard Normal Cumulative Probability Table.)

.0029

A population has a mean of 53 and a standard deviation of 21. A sample of 49 observations will be taken. The probability that the sample mean will be greater than 57.95 is

.0495 (57.95-53) / (21/ square root of 49) = 1.65 Pr (z<1.65) = .9505 1-.9505= .0495

A population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. The probability that the mean from that sample will be between 183 and 186 is

.1359 183-180/(24/sqrt (64)) =1 Pr (z<1) = 0.8413 186-180/(24/sqrt(64)) Pr (z<2) =.9772 .9772-.8413

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. Refer to Exhibit 8-1. The standard error of the mean is

.29

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. Refer to Exhibit 8-1. With a 0.95 probability, the margin of error is approximately

.39 =1.96 * 1.8 / Square root (81)

A population of size 1,000 has a proportion of 0.5. Therefore, the proportion and the standard deviation of the sample proportion for samples of size 100 are

.5 and .050 Square root (0.5*0.5/100)

In a random sample of 144 observations, = 0.6. The 95% confidence interval for P is

.52 to .68 0.6 + 1.96 * square rt ((0.6) * (0.4) / 144)

Four hundred registered voters were randomly selected asked whether gun laws should be changed. Three hundred said "yes," and one hundred said "no." Refer to Exhibit 7-2. The point estimate of the proportion in the population who will respond "yes" is

.75

If an interval estimate is said to be constructed at the 90% confidence level, the confidence coefficient would be

.9

X is a normally distributed random variable with a mean of 8 and a standard deviation of 4. The probability that x is between 1.48 and 15.56 is (Use the Standard Normal Cumulative Probability Table.)

.9190

X is a normally distributed random variable with a mean of 12 and a standard deviation of 3. The probability that x equals 19.62 is (Use the Standard Normal Cumulative Probability Table.) Select one:

0.000

X is a normally distributed random variable with a mean of 22 and a standard deviation of 5. The probability that x is less than 9.7 is (Use the Standard Normal Cumulative Probability Table.)

0.0069 9.7-22/5= -2.46 z-score=0.0069

A population has a mean of 80 and a standard deviation of 7. A sample of 49 observations will be taken. The probability that the mean from that sample will be larger than 82 is

0.0228 (82-80) / (7 / square root of 49) = 2 Pr (z<2) = 0.9772 1-0.9772=0.0228

A random sample of 150 people was taken from a very large population. Ninety of the people in the sample were females. The standard error of the proportion of females is

0.0400

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the probability that a randomly selected item will weigh between 11 and 12 ounces?

0.0440 11: 11-8/2=1.5 z-score 1.5= 0.9332 12: 12-8/2=2 z-score 2= 0.9772 0.9772-0.9332 = 0.0440

A sample of 400 observations will be taken from a process (an infinite population). The population proportion equals 0.8. The probability that the sample proportion will be greater than 0.83 is

0.0668 (.83-.8) / square root (0.8 * (1-0.8) / 400) =1.5 z-score=.9332, the area to the right is 1-.9332=0.0668

In a sample of 400 voters, 360 indicated they favor the incumbent governor. The 95% confidence interval of voters not favoring the incumbent is

0.071 to 0.129 =0.1 + 1.96 * Square root ((0.1) * (0.9)/400)

A sample of 51 observations will be taken from a process (an infinite population). The population proportion equals 0.85. The probability that the sample proportion will be between 0.9115 and 0.946 is

0.0819

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the probability that a randomly selected item will weigh more than 10 ounces?

0.1587 10-8/2 z-score= 1 --> .8413 1-.8413=.1587

A random sample of 1000 people was taken. Four hundred fifty of the people in the sample favored Candidate A. The 95% confidence interval for the true proportion of people who favors Candidate A is

0.419 to 0.481 0.45+1.96*square root ((0.45) *(1-0.45)/1000)

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What percentage of items will weigh between 6.4 and 8.9 ounces?

0.4617

Given that z is a standard normal random variable, what is the value of z if the area to the right of z is 0.1401? Use the Standard Normal Cumulative Probability Table.

1.08

Given that z is a standard normal random variable, what is the value of z if the area to the right of z is 0.1112? Use the Standard Normal Cumulative Probability Table.

1.22

A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. Refer to Exhibit 8-6. The value of the margin of error at 95% confidence is

1.611 1.993 * 7 / Square root (75)

A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. Refer to Exhibit 8-6. If we want to determine a 95% confidence interval for the average hourly income, the value of "t" statistics is

1.993

A sample of 225 elements from a population with a standard deviation of 75 is selected. The sample mean is 180. The 95% confidence interval for μ is

170.2 to 189.8 180 + 1.96 * 75 / square root (225)

Imagine we are trying to sell to a customer who demands that the mean of a random sample of 64 bulbs lasts at least 2,050 hours before they will buy. The population mean = 2,000 hours, and the population standard deviation is 100 hours. What mean length of bulb life could you be 90% confident that the sample mean will be at least that long? (Use the Standard Normal Cumulative Probability Table. Round to nearest integer)

1984 10% to the left z-value=-1.285 Mean=2000 St. Dev. = 64 value=mean+std.dev*z-value

A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is

24.4 to 25.6 =25 + 2.387 * 2 / Square root (64)

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What percentage of items will weigh at least 11.7 ounces?

3.22% 11.8-8/2 = 1.85 z-score 1.85 --> .9678 1-9678=.0322

Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 20 gallons and a standard deviation of 8 gallons. If the manager of Pep Zone wants the probability of a stockout during replenishment lead-time to be no more than .025, what should the reorder point be? (Use the Standard Normal Cumulative Probability Table, just enter the number without any units. Keep two decimal places.)

35.68, 2.5% to the left z-value=1.96 =20+1.96*8 =35.68

Random samples of size 17 are taken from a population that has 200 elements, a mean of 36, and a standard deviation of 8. Refer to Exhibit 7-5. The mean and the standard deviation of the sampling distribution of the sample means are (Hint: this is a finite population question, we need the finite population correction factor, see page 21 of the ppt file)

36 and 1.86 Standard deviation- Square root (200-17/200-1)*(8/square root(17) NOT ON EXAM

A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. Refer to Exhibit 8-6. The 95% confidence interval for the average hourly wage of all information system managers is

39.14 to 42.36

In order to determine an interval for the mean of a population with unknown standard deviation a sample of 61 items is selected. The mean of the sample is determined to be 23. The number of degrees of freedom for reading the t value is

60

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. Refer to Exhibit 8-1. If the sample mean is 9 hours, then the 95% confidence interval is

8.61 t0 9.39 hours

A continuous random variable may assume

All values in an interval or collection of intervals

A sample of 25 observations is taken from a process (an infinite population). The sampling distribution of is

Approximately normal if np >_ 5 and n(1-p) >_5

A 95% confidence interval for a population mean is determined to be 100 to 120. If the confidence coefficient is reduced to 0.90, the interval for μ

Becomes narrower

As the number of degrees of freedom for a t distribution increases, the difference between the t distribution and the standard normal distribution

Becomes smaller

The fact that the sampling distribution of the sample mean can be approximated by a normal probability distribution whenever the sample size is large is based on the

Central limit theorem

The following data was collected from a simple random sample from a process (an infinite population). 13 15 14 16 12 Refer to Exhibit 7-1. The mean of the population

Could be any value

A normal probability distribution

Is a continuous probability distribution

The sampling distribution of the sample mean

Is the probability distribution showing all possible values of the sample mean

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the probability that a randomly selected item weighs exactly 8 ounces? Use the Standard Normal Cumulative Probability Table.

None of the alternative answers are correct

The standard deviation of all possible values is called the

Standard error of the mean

As the sample size increases, the

Standard error of the mean decreases

Excel's NORM.INV function can be used to compute

The normally distributed x value given a cumulative probability

In interval estimation, the t distribution is applicable only when

The sample standard deviation is used to estimate the population standard deviation

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. What is the random variable in this experiment?

The weight of items produced by a machine

__________ is a property of a point estimator that is present when the expected​ value of the point estimator is equal to the population parameter it estimates.

Unbiased

A standard normal distribution is a normal distribution with

a mean of 0 and a standard deviation of 1

A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is 15. a. With 95% confidence, what is the margin of error? (Specify to 2 decimal places. Round the final answer, no rounding on intermediate steps) b. Compute the 95% confidence interval for the population mean (Specify to 2 decimal places). c. Assume that the same sample mean was obtained from a sample of 120 items. With 95% confidence, what is the margin of error? (Specify to 2 decimal places. Round the final answer, no rounding on intermediate steps) d. Compute the 95% confidence interval for the population mean (Specify to 2 decimal places).

a. 3.80 1.96*15/Square root (60) b. 76.20, 83.80 80+1.96*15/Square root (60) c. 2.68 1.96 * 15 / Square root (120) d. 80+1.96*15/Square root (120)

Using an α = 0.04 a confidence interval for a population proportion is determined to be 0.65 to 0.75. If the level of significance is decreased, the interval for the population proportion (Hint: check textbook section 8.1 for the definition of level of significance)

becomes wider the level of significance is alpha, smaller alpha has bigger confidence coefficient

Excel's NORM.DIST function can be used to compute

cumulative probabilities for a normally distributed x value

The following data was collected from a simple random sample from a process (an infinite population). 13 15 14 16 12 Refer to Exhibit 7-1. The point estimate of the population standard deviation is (Hint: page 16 of ppt file)

is 1.581

The following data was collected from a simple random sample from a process (an infinite population). 13 15 14 16 12 Refer to Exhibit 7-1. The point estimate of the population mean (Hint: page 16 of ppt file

is 14

The standard deviation of p (line over it) is referred to as the

standard error of the proportion

If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect

the size of the confidence interval to increase

"DRUGS R US" is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that the bottles of vitamins marked 6 ounces vary in content with a standard deviation of 0.3 ounces. Assume the contents of the bottles are normally distributed. Ninety-five percent of the bottles will contain at least how many ounces? (Use the Standard Normal Cumulative Probability Table. Check page 32 on ppt file for an example. Just enter the number without any units. Keep four decimal places.)

z-value: -1.645 5% to the left =6-1.645*0.3