Tom 302

Front 1

Standard Normal Distribution

Back 1

Allows us to use a single table to describe areas beneath the curve

Front 2

Standard Error of the Mean

Back 2

Standard deviation of the sampling distribution of the mean

Front 3

Sampling Distribution of the Mean

Back 3

The probability distribution of the sample means for all possible samples of that particular size

Front 4

Point Estimate

Back 4

a single number that estimates the exact value of the population parameter of interest

Front 5

Interval Estimate

Back 5

includes a range of possible values that are likely to include the actual population paramater

Front 6

For two samples from what are assumed to be normally distributed populations, the sample sizes and standard deviations are n1= 28, s1=103.1, n2=41, and s2=133.5. At the 0.10 level of significance, test the null hypothesis that the population variances are equal. Would your conclusion be different if the test had been conducted at the 0.05 level? At the 0.02 level?

Back 6

Comparing variances from two independent samples
• F-test, with test statistic:

Front 7

During the course of a year, 38.0% of the 213 merchant ships lost were general cargo carriers, while 43.8% of the 219 ships lost during the following year were in this category. Using a two-tail test at the 0.05 level, examine whether this difference could have been the result of chance variation from one year to the next.

Back 7

Comparing proportions from two independent samples
• z-test, with test statistic
1. When the hypothesized difference is zero (the usual case):

Front 8

In examining the ability of users to complete a variety of information-seeking tasks on their mobile devices, Nielsen Norman Group assigned sample members to get the answers to a variety of questions like, “How many calories are there typically in a slice of thincrust pizza?” The success rate for achieving such tasks was 75% for those using touch-screen phones like the iPhone to access the mobile Internet, compared to 80% for persons accessing websites on a conventional personal computer. Assuming these percentages to be based on independent samples of 500 persons each, and using the 0.01 level of significance, can we conclude that the population success rate with a conventional PC is greater than that when using the mobile Internet and touchscreen phone?

Back 8

Comparing proportions from two independent samples
• z-test, with test statistic
1. When the hypothesized difference is zero (the usual case):

Front 9

According to the U.S. Bureau of Labor Statistics, personal expenditures for entertainment fees and admissions averaged $349 per person in the Northeast and $420 in the West. Assuming that these data involved (1) sample sizes of 800 and 600 and (2) standard deviations of $215 and $325, respectively, use a z-test and the 0.01 level of significance in testing the difference between these means. Determine and interpret the p-value for the test. Construct and interpret the 99% confidence interval for the test.

Back 9

z -test approximation for comparing the means of two independent samples, o1 and o2 unknown, and each n >= 30

Front 10

A tire company is considering switching to a new type of adhesive designed to improve tire reliability in high-temperature and overload conditions. In laboratory “torture” tests with temperatures and loads 90% higher than the maximum normally encountered in the field, 15 tires constructed with the new adhesive run an average of 65 miles before failure, with a standard deviation of 14 miles. For 18 tires constructed with the conventional adhesive, the mean mileage before failure was 53 miles, with a standard deviation of 22 miles. Assuming normal populations and using the 0.05 level of significance, can we conclude that the new adhesive is superior to the old under such test conditions? What is the most accurate statement that could be made about the p-value for this test?

Back 10

Unequal-variances t-test for comparing the means of two independent samples, o1 and o2 unknown and not assumed to be equal

Front 11

According to the human resources director of a plant, no more than 5% of employees hired in the past year have violated their preemployment agreement not to use any of five illegal drugs. The agreement specified that random urine checks could be carried out to ascertain compliance. In a random sample of 400 employees, screening detected at least one of these drugs in the systems of 8% of those tested. At the 0.025 level, is the human resources director’s claim credible? Determine and interpret the p-value for the test.

Back 11

Testing a Proportion
z-test, with test statistic:

Front 12

A maintenance supervisor is comparing the standard version of an instructional booklet with one that has been claimed to be superior. An experiment is conducted in which 26 technicians are divided into two groups, provided with one of the booklets, then given a test a week later. For the 13 using the standard version, the average exam score was 72.0, with a standard deviation of 9.3. For the 13 given the new version, the average score was 80.2, with a standard deviation of 10.1. Assuming normal populations with equal standard deviations, and using the 0.05 level of significance, does the new booklet appear to be better than the standard version?

Back 12

Pooled-variances t-test for comparing the means of two independent samples, o1 and  o2 unknown and assumed to be equal

Front 13

A scrap metal dealer claims that the mean of his cash sales is “no more than $80,” but an Internal Revenue Service agent believes the dealer is untruthful. Observing a sample of 20 cash customers, the agent finds the mean purchase to be $91, with a standard deviation of $21. Assuming the population is approximately normally distributed, and using the 0.05 level of significance, is the agent’s suspicion confirmed?

Back 13

Testing a Mean, Population Standard Deviation Unknown
t-test, with test statistic:

Front 14

In the past, patrons of a cinema complex have spent an average of $5.00 for popcorn and other snacks, with a standard deviation of $1.80. The amounts of these expenditures have been normally distributed. Following an intensive publicity campaign by a local medical society, the mean expenditure for a sample of 18 patrons is found to be $4.20. In a one-tail test at the 0.05 level of significance, does this recent experience suggest a decline in spending? Determine and interpret the p-value for the test.

Back 14

Testing a Mean, Population Standard Deviation Known
z-test, with test statistic:

Front 15

Based on a pilot study, the population standard deviation of scores for U.S. high school graduates taking a new version of an aptitude test has been estimated as 3.7 points. If a larger study is to be undertaken, how large a simple random sample will be necessary to have 99% confidence that the sample mean will not differ from the actual population mean by more than 1.0 point?

Back 15

Required Sample Size for Estimating a Population Proportion

Front 16

A pharmaceutical company found that 46% of 1000U.S. adults surveyed knew neither their blood pressure nor their cholesterol level. Assuming the persons surveyed to be a simple random sample of U.S. adults, construct a 95% confidence interval for o= the population proportion of U.S. adults who would have given the same answer if a census had been taken instead of a survey.

Back 16

Confidence interval limits for the population proportion:

Front 17

The service manager of Appliance Universe has recorded the times for a simple random sample of 50 refrigerator service calls taken from last year’s service records. The sample mean and standard deviation were 25 minutes and 10 minutes, respectively.
a. Construct and interpret the 95% confidence interval for the mean.
b. It’s quite possible that the population of such times is strongly skewed in the positive direction—that is, some jobs, such as compressor replacement, might take 3 or 4 hours. If this were true, would the interval constructed in part (a) still be appropriate? Explain your answer.

Back 17

Confidence interval limits for the population mean,  o unknown:

Front 18

An assembly process includes a torque wrench device that automatically tightens compressor housing bolts; the device has a known process standard deviation of o = 3 lb-ft in the torque applied. A simple random sample of 35 nuts is selected, and the average torque to which they have been tightened is 150 lb-ft. What is the 95% confidence interval for the average torque being applied during the assembly process?

Back 18

Confidence interval limits for the population mean,  o known:

Front 19

A civic organization includes 200 members, who have an average income of $58,000, with a standard deviation of $10,000. A simple random sample of n = 30 members is selected to participate in the annual fund-raising drive. What is the probability that the average income of the fund-raising group will be at least $60,000?

Back 19

Standard error for the sample mean when sampling without replacement
from a finite population:

Front 20

The average length of a hospital stay in general or community hospitals in the United States is 5.5 days. Assuming a population standard deviation of 2.5 days and a simple random sample of 50 patients, what is the probability that the average length of stay for this group of patients will be no more than 6.5 days? If the sample size had been 8 patients instead of 50, what further assumption(s) would have been necessary in order to solve this problem?

Back 20

z-score for the sampling distribution of the mean, normally distributed
population: