Mar 26, 2017

- Solve the problem.
**1.**Find the value of –z_{α}_{/2 }that corresponds to a level of confidence of 94.76 percent.- –1.62
- 1.94
- –1.94
- 0.0262

**2.**The following confidence interval is obtained for a population proportion, p:

0.620 < p < 0.658.

Use these confidence interval limit to find the margin of error, E.- 0.019
- 0.038
- 0.639
- 0.020

- Find the margin of error for the 95% confidence interval used to estimate the population proportion.
**3.**In a survey of 7100 TV viewers, 40% said they watch network news programs.- 0.0131
- 0.00855
- 0.0114
- 0.0150

- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
**4.**n = 87, x = 48; 98 percent- 0.448 < p < 0.656
- 0.427 < p < 0.677
- 0.447 < p < 0.657
- 0.428 < p < 0.676

- Find the minimum sample size you should use to assure that your estimate of p̂ will be within the required margin of error the population p.
**5.**Margin of error: 0.006; confidence level: 95%, p̂ and q̂ are unknown.- 38,416
- 16,112
- 26,678
- 38,415

**6.**Margin of error: 0.04; confidence level: 90%; from a prior study, p̂ is estimated by 0.28.- 303
- 14
- 1023
- 341

- Solve the problem.
**7.**50 people are selected randomly from a certain population and it is found that 14 people in the sample are over 6 feet tall. What is the point estimate of the true proportion of people in the population who are over 6 feet tall?- 0.20
- 0.50
- 0.72
- 0.28

- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
**8.**A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval.- 0.471 < p < 0.472
- 0.435 < p < 0.508
- 0.444 < p < 0.500
- 0.438 < p < 0.505

- Solve the problem.
**9.**Suppose that n trials of a binomial experiment result in no successes. According to the “Rule of Three”, we have 95% confidence interval for the true population proportion has an upper bound of 3/n. If a manufacturer randomly selects 25 computers for quality control and finds no defective computers, what statement can you make by using the rule of three, about the proportion p, of all its computers which are defective?- We are 95% confident that p is greater than 3/25.
- The value of p cannot be greater than 3/25.
- We are 95% confident that p lies between 1/25 and 3/25.
- We are 95% confident that p does not exceed 3/25.

- Determine whether the given conditions justify using the margin of error E = z
_{α}_{/}_{2}**10.**The sample size is n = 7, σ = 12.2, and the original population is normally distributed.- No
- Yes

- Use the confidence level and sample data to find the margin of error E.
**11.**Weights of eggs: 95% confidence; n = 46, x̅ = 1.68 oz, σ = 0.46 oz.- 0.13 oz
- 0.11 oz
- 6.78 oz
- 0.02 oz

- Use the confidence level and sample data to find a confidence interval for estimation the population μ.
**12.**A random sample of 94 light bulbs had a mean life of x̅ = 587 hours with a standard deviation of σ = 36 hours. Construct a 90 percent confidence interval for the mean life, μ, of all light bulbs of this type.- 578 < μ < 596
- 580 < μ < 594
- 581 < μ < 593
- 577 < μ < 597

- Use the margin of error, confidence level, and standard deviation σ to find the minimum sample size required to estimate an unknown population mean μ.
**13.**Margin of error: $133, confidence level: 95%, σ = $530.- 50
- 62
- 20
- 54

- Do one of the following, as appropriate:

(a) Find the critical value z_{α}_{/2};

(b) Find the critical value t_{α}_{/2};

(c) state that neither the normal nor the t distribution applies.**14.**95%; n = 11; σ is known; population appears to be very skewed.- z
_{α}_{/2}= 1.812 - z
_{α}_{/2}= 1.96 - Neither the normal nor the t distribution applies.
- t
_{α}_{/2}_{ }= 2.228

- z

- Find the margin of error.
**15.**95% confidence interval; n = 91; x̅ = 28; s = 14.1.- 2.93
- 4.25
- 2.05
- 1.85

- Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
**16.**The principal randomly selected six students to take an aptitude test. Their scores were:

83.0 84.1 83.5 83.7 84.1 73.5.

Determine a 90 percent confidence interval for the mean score for all students.- 85.52 < μ < 78.45
- 85.42 < μ < 78.55
- 78.55 < μ < 85.42
- 78.45 < μ < 85.52

- Solve the problem.
**17.**Find the chi-square value X^{2}_{L}corresponding to a sample size of 20 and a confidence level of 99 percent.- 36.191
- 6.844
- 38.582
- 7.633

- Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution.
**18.**A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the standard deviation, σ, of the scores of all subjects.- 16.9 < σ < 29.3
- 16.6 < σ < 28.6
- 17.2 < σ < 27.2
- 17.5 < σ < 27.8

- Find the appropriate minimum sample size.
**19.**You want to be 99% confident that the sample standard deviation s is within 5% of the population standard deviation.- 1,335
- 2,638
- 2,434
- 923

- Use the give degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution.
**20.**The daily intakes of milk (in ounces) for ten randomly selected people were:

22.3 12.9 23.1 26.3 10.8

26.8 27.4 29.7 19.0 19.9

Find a 99 percent confidence interval for the population standard deviation σ.- (3.74, 12.81)
- (0.89, 3.40)
- (3.87, 14.26)
- (3.87, 12.81)

*Problems with solutions in Elementary Statistics — Estimating a Population Proportion, Estimating a Population Mean: σ Known, Estimating a Population Mean: σ Not Known, Estimating a Population Variance.***Directions:**

*Choose the correct answer for multiple-choice items.*

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