Exam 9: Confidence Intervals

arrow
  • Select Tags
search iconSearch Question
flashcardsStudy Flashcards
  • Select Tags

An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas. They test a sample of five cars and obtain a mean of 28.2 miles. How many cars should the environmental group test if they wish to estimate μ, mean miles per 1 gallon, to within .5 miles and be 99 percent confident? Assume a population standard deviation of 2.7 miles.

(Multiple Choice)
4.9/5
(87)

In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Past data indicate that the standard deviation is .25 inches. How many manufactured bolts should be sampled in order to make us 95 percent confident that the sample mean bolt length is within .02 inches of the true mean bolt length?

(Multiple Choice)
5.0/5
(76)

The 99 percent confidence interval for the average weight of a product is from 71.36 lb to 78.64 lb. Is μ = 71 a possible population mean using that 99 percent confidence interval?

(Multiple Choice)
4.9/5
(78)

When sample size is 11, find t.001.

(Short Answer)
4.9/5
(65)

As the sample size n increases, the width of the confidence interval ________.

(Multiple Choice)
4.7/5
(83)

When sample size is 20, find t.001.

(Short Answer)
4.8/5
(72)

Find a 98 percent confidence interval for p when Find a 98 percent confidence interval for p when   = .25 and n = 400. = .25 and n = 400.

(Multiple Choice)
4.8/5
(76)

A sample set of weights in pounds are 1.01, .95, 1.03, 1.04, .97, .97, .99, 1.01, and 1.03. Assume the population of weights is normally distributed. Find a 99 percent confidence interval for the mean population weight.

(Multiple Choice)
4.7/5
(79)

A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm, respectively. Find the 95 percent confidence interval for the mean height of seedlings in the population from which the sample was selected.

(Multiple Choice)
4.9/5
(86)

When determining the sample size, n, if the value found for n is 79.2, we would choose to sample 79 observations.

(True/False)
5.0/5
(87)

In 2005, 13,000 Internet users were surveyed and asked about their willingness to pay fees for access to websites. Of these, 2,938 were definitely not willing to pay such fees. How large a sample is necessary to estimate the proportion of interest to within 2 percent in a 95 percent confidence interval?

(Multiple Choice)
4.8/5
(68)

When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on a sample of n = 100 will be narrower than a confidence interval for a population mean based on a sample of n = 50.

(True/False)
4.9/5
(81)

A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of .996 ounces with a standard deviation of .004 ounces. How many pieces of candy must we sample if we want to be 99 percent confident that the sample mean is within .001 ounces of the true mean?

(Multiple Choice)
4.8/5
(75)

A random sample n = 15 is taken from a population assumed to be normal, and A random sample n = 15 is taken from a population assumed to be normal, and   = 1.2 and the sample variance is .36. Calculate a 99 percent confidence interval for μ. = 1.2 and the sample variance is .36. Calculate a 99 percent confidence interval for μ.

(Multiple Choice)
4.7/5
(89)

In a study of 265 engineering students, the average score on the final exam in the statistics course was 63.8 and s = 3.08. What is a 95 percent confidence for μ?

(Multiple Choice)
4.8/5
(73)

The t distribution approaches the ________ distribution as the sample size ________.

(Multiple Choice)
4.8/5
(72)

The standard error of the sample mean is σ/ The standard error of the sample mean is σ/   . .

(True/False)
5.0/5
(81)

When sample size is 16, find t.001.

(Short Answer)
4.9/5
(77)

An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered by the company. A random sample of 240 claims shows that the insurance company covered 90 accident claims, while 150 claims were not covered. Use a confidence interval of 95 percent and determine the margin of error.

(Multiple Choice)
4.8/5
(81)

When the sample size and sample standard deviation remain the same, a 99 percent confidence interval for a population mean, μ, will be narrower than the 95 percent confidence interval for μ.

(True/False)
5.0/5
(72)
Showing 121 - 140 of 149
close modal

Filters

  • Essay(0)
  • Multiple Choice(0)
  • Short Answer(0)
  • True False(0)
  • Matching(0)