Mar 31, 2017

- Solve the problem.
**1.**Find the critical value of z_{α}_{/2 }that corresponds to a degree of confidence of 98%.- 2.05
- 1.75
- 2.33
- 2.575

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

(0.870, 0.897)

Use these confidence interval limit to find the point estimate, p̂.- 0.882
- 0.885
- 0.870
- 0.894

- Find the margin of error for the 95% confidence interval used to estimate the population proportion.
**3.**In a clinical test with 2440 subjects, 70% showed improvement from the treatment.- 0.0196
- 0.0236
- 0.0182
- 0.0175

- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
**4.**n = 107, x = 66; 88 percent- 0.539 < p < 0.695
- 0.540 < p < 0.694
- 0.543 < p < 0.691
- 0.544 < p < 0.690

- 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: 0.07; confidence level: 97%, p̂ and q̂ are unknown.- 240
- 241
- 111
- 112

**6.**Margin of error: 0.008; confidence level: 99%; from a prior study, p̂ is estimated by 0.164.- 12,785
- 8230
- 14,205
- 1145

- Solve the problem.
**7.**38 randomly picked people were asked if they rented or owned their own home, 11 said they rented. Obtain a point estimate of the true proportion of home owners.- 0.737
- 0.289
- 0.711
- 0.224

- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
**8.**A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate.- 0.308 < p < 0.438
- 0.304 < p < 0.442
- 0.316 < p < 0.430
- 0.301 < p < 0.445

- Solve the problem.
**9.**A one-sided confidence interval for p can be written as p < p̂ + E or p > p̂ - E where the margin of error E is modified by replacing z_{α}_{/2}_{ }with_{ }z_{α}. If a teacher wants to report that the fail rate on a test is at most x with 90% confidence, construct the appropriate one-sided confidence interval. Assume that a simple random sample of 58 students results in 6 who fail the test.- p > 0.155
- p > 0.052
- p < 0.052
- p < 0.155

- Determine whether the given conditions justify using the margin of error E = z
_{α}_{/2}**10.**The sample size is n = 8, and σ is not known.- Yes
- No

- Use the confidence level and sample data to find the margin of error E.
**11.**College students’ annual earnings: 99% confidence; n = 66, x̅ = $3903, σ = $801.- $254
- $8
- $230
- $1237

- Use the confidence level and sample data to find a confidence interval for estimation the population μ.
**12.**A random sample of 100 full-grown lobsters had a mean weight of 22 ounces and a standard deviation of 3.7 ounces. Construct a 98 percent confidence interval for the population mean μ.- 22 < μ < 24
- 21 < μ < 23
- 20 < μ < 22
- 21 < μ < 24

- 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: $136, confidence level: 99%, σ = $503.- 91
- 53
- 10
- 46

- 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.**90%; n = 9; σ = 4.2; population appears to be very skewed.- z
_{α}_{/2}= 2.896 - Neither the normal nor the t distribution applies.
- z
_{α}_{/2}= 2.306 - z
_{α}_{/2 }= 2.365

- z

- Find the margin of error.
**15.**95% confidence interval; n = 12; x̅ = 35.6; s = 6.4.- 4.066
- 3.659
- 3.050
- 4.879

- 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 amount (in ounces) of juice in eight randomly selected juice bottles are:

15.1 15.6 15.6 15.9

15.9 15.2 15.1 15.2

Construct a 98 percent confidence interval for the mean amount of juice in all such bottles.- 15.77 < μ < 15.13
- 15.09 < μ < 15.81
- 15.03 < μ < 15.87
- 15.87 < μ < 15.03

- Solve the problem.
**17.**Find the critical value X^{2}_{L}corresponding to a sample size of 3 and a confidence level of 90 percent.- 5.991
- 9.21
- 0.103
- 0.0201

- 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.**Weights of men: 90% confidence; n = 14, x̅ = 156.9 lb, s = 11.1 lb.- 9.0 lb < σ < 2.7 lb
- 8.5 lb < σ < 16.5 lb
- 8.7 lb < σ < 14.3 lb
- 8.2 lb < σ < 15.6 lb

- Find the appropriate minimum sample size.
**19.**To be able to say with 95% confidence level that the standard deviation of a data set is within 10% of the population’s standard deviation, the number of observations within the data set must be greater than or equal to what quantity?- 191
- 805
- 335
- 250

- 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 amounts (in ounces) of juice in eight randomly selected juice bottles are:

15.8 15.7 15.0 15.7

15.2 15.2 15.5 15.0

Find a 98 percent confidence interval for the population standard deviation σ.- (0.20, 0.78)
- (0.20, 0.67)
- (0.19, 0.67)
- (0.22, 0.84)

*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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