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Chapter 7 Exam Review. Solve the problem. Essay

1) Find the critical value that corresponds to a degree of confidence of 91%. A) 1.70B) 1.34 C) 1.645 D) 1.75

2) The following confidence interval is obtained for a population proportion, p:0.817 < p < 0.855 Use these confidence interval limits to find the point estimate, A) 0.839 B) 0.836 C) 0.817 D) 0.833

Find the margin of error for the 95% confidence interval used to estimate the population proportion. 3) n = 186, x = 103
A) 0.0643 B) 0.125 C) 0.00260 D) 0.0714

Find the minimum sample size you should use to assure that your estimate of will be within the required margin of error around the population p. 4) Margin of error: 0.002; confidence level: 93%; and unknown A) 204,757 B) 410 C) 204,750 D) 405

5) Margin of error: 0.07; confidence level: 95%; from a prior study, is estimated by the decimal equivalent of 92%.
A) 58 B) 174 C) 51 D) 4

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
6) When 343 college students are randomly selected and surveyed, it is found that 110 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. A) 0.256 < p < 0.386 B) 0.279 < p < 0.362C) 0.271 < p < 0.370 D) 0.262 < p < 0.379

Determine whether the given conditions justify using the margin of error E = when finding a confidence interval estimate of the population mean . 7) The sample size is n = 9, is not known, and the original population is normally distributed. A) Yes B) No

Use the confidence level and sample data to find the margin of error E. 8) Systolic blood pressures for women aged 18-24: 94% confidence; n = 92, x = 114.9 mm Hg, = 13.2 mm Hg
A) 47.6 mm Hg B) 2.3 mm Hg C) 2.6 mm Hg D) 9.6 mm Hg

Use the confidence level and sample data to find a confidence interval for estimating the population .
9) A group of 52 randomly selected students have a mean score of 20.2 with a standard deviation of 4.6 on a placement test. What is the 90 percent confidence interval for the mean score, , of all students taking the test?

A) 19.1 < < 21.3 B) 18.7 < < 21.7C) 19.0 < < 21.5 D) 18.6 < < 21.8

Use the margin of error, confidence level, and standard deviation to find the minimum sample size required to estimate an unknown population mean . 10) Margin of error: $100, confidence level: 95%, = $403

A) 91 B) 63 C) 108 D) 44

Formula sheet for Final Exam

Mean Standard deviation Variance =

Mean from a frequency distribution Range rule of thumb

Empirical Rule 68-95-99.7 z – score weighted mean


if A and B are mutually exclusive
if A and B are not mutually exclusive
if A and B are independent
if A and B are dependent
Complementary events

mean of a probability distribution
standard deviation of a
probability distribution
Binomial probability
Binomial probability calculator
Exactly binompdf(n,p,x)
At least 1 – binomcdf(n,p,x –1)
At most binomcdf(n,p,x)
Binomial mean
Binomial standard deviation
Expected value

Margin of error p
Sample size p or
Margin of error mean
Sample size mean
Margin of error mean

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