Question

multiple choice. choose the one alternative that best completes the statement or answers the question.
find the indicated critical z value.

1. find the critical value z_α/2 that corresponds to a 91% confidence level.

a) 1.75 b) 1.34 c) 1.70 d) 1.645

1. find the value of z_α/2 that corresponds to a confidence level of 89.48%.

a) 0.0526 b) 1.25 c) -1.62 d) 1.62
express the confidence interval using the indicated format.

1. express the confidence interval 0.38 < p < 0.54 in the form of \\(\hat{p}\pm e\\).

a) 0.38 ± 0.08 b) 0.46 ± 0.08 c) 0.46 ± 0.16 d) 0.38 ± 0.16

1. express the confidence interval (0.432, 0.52) in the form of \\(\hat{p}\pm e\\).

a) 0.432 ± 0.088 b) 0.476 ± 0.088 c) 0.432 ± 0.044 d) 0.476 ± 0.044
solve the problem.

1. the following confidence interval is obtained for a population proportion, p: 0.753 < p < 0.797. use these confidence interval limits to find the point estimate, \\(\hat{p}\\).

a) 0.753 b) 0.770 c) 0.775 d) 0.780

1. the following confidence interval is obtained for a population proportion, p: (0.399, 0.437). use these confidence interval limits to find the margin of error, e.

a) 0.038 b) 0.019 c) 0.020 d) 0.017
assume that a sample is used to estimate a population proportion p. find the margin of error e that corresponds to the given statistics and confidence level. round the margin of error to four decimal places.

1. 90% confidence; n = 300, x = 140

a) 0.0509 b) 0.0593 c) 0.0565 d) 0.0474

1. 95% confidence; n = 2388, x = 1672

a) 0.0156 b) 0.0206 c) 0.0184 d) 0.0248

1. 98% confidence; the sample size is 800, of which 40% are successes

a) 0.0446 b) 0.0404 c) 0.0339 d) 0.0355

1. 90% confidence; the sample size is 1510, of which 35% are successes

a) 0.0157 b) 0.0252 c) 0.0241 d) 0.0202
use the given data to find the minimum sample size required to estimate the population proportion.

1. margin of error: 0.012; confidence level: 93%; \\(\hat{p}\\) and \\(\hat{q}\\) unknown

a) 5537 b) 5688 c) 7687 d) 4685

Explanation:

Step1: Recall confidence - level formula

For a confidence level $C$, $\alpha=1 - C$. Then $z_{\alpha/2}$ is the $z$ - value such that the area in the two - tails of the standard normal distribution is $\alpha$.

Step2: Calculate $\alpha$ for 91% confidence level

For $C = 0.91$, $\alpha=1 - 0.91=0.09$, and $\alpha/2=0.045$. Looking up in the standard normal table, $z_{\alpha/2}=1.695\approx1.70$. So the answer to 1) is C.

Step3: Calculate $\alpha$ for 89.48% confidence level

For $C = 0.8948$, $\alpha=1 - 0.8948 = 0.1052$, and $\alpha/2=0.0526$. Looking up in the standard normal table, $z_{\alpha/2}=1.62$. So the answer to 2) is D.

Step4: Recall confidence - interval formula for proportion

The confidence interval for a proportion $p$ is given by $\hat{p}-E\lt p\lt\hat{p} + E$. The mid - point of the interval $a\lt p\lt b$ is $\hat{p}=\frac{a + b}{2}$, and $E=\frac{b - a}{2}$.

Step5: Solve for 3)

For the interval $0.38\lt p\lt0.54$, $\hat{p}=\frac{0.38 + 0.54}{2}=0.46$ and $E=\frac{0.54 - 0.38}{2}=0.08$. So the answer is B.

Step6: Solve for 4)

For the interval $(0.432,0.52)$, $\hat{p}=\frac{0.432+0.52}{2}=0.476$ and $E=\frac{0.52 - 0.432}{2}=0.044$. So the answer is D.

Step7: Recall point - estimate formula

The point estimate $\hat{p}$ of a population proportion $p$ for a confidence interval $a\lt p\lt b$ is $\hat{p}=\frac{a + b}{2}$.

Step8: Solve for 5)

For the interval $0.753\lt p\lt0.797$, $\hat{p}=\frac{0.753 + 0.797}{2}=0.775$. So the answer is C.

Step9: Recall margin - of - error formula

The margin of error $E$ for a confidence interval $a\lt p\lt b$ is $E=\frac{b - a}{2}$.

Step10: Solve for 6)

For the interval $(0.399,0.437)$, $E=\frac{0.437 - 0.399}{2}=0.019$. So the answer is B.

Step11: Recall margin - of - error formula for proportion

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}=\frac{x}{n}$ and for a confidence level $C$, $\alpha = 1 - C$.

Step12: Solve for 7)

For $C = 0.90$, $\alpha=0.10$, $\alpha/2 = 0.05$, $z_{\alpha/2}=1.645$, $n = 300$, $x = 140$, $\hat{p}=\frac{140}{300}\approx0.4667$. Then $E=1.645\sqrt{\frac{0.4667\times(1 - 0.4667)}{300}}\approx0.0474$. So the answer is D.

Step13: Solve for 8)

For $C = 0.95$, $\alpha=0.05$, $\alpha/2=0.025$, $z_{\alpha/2}=1.96$, $n = 2388$, $x = 1672$, $\hat{p}=\frac{1672}{2388}\approx0.6927$. Then $E=1.96\sqrt{\frac{0.6927\times(1 - 0.6927)}{2388}}\approx0.0184$. So the answer is C.

Step14: Solve for 9)

For $C = 0.98$, $\alpha=0.02$, $\alpha/2=0.01$, $z_{\alpha/2}=2.33$, $n = 800$, $\hat{p}=0.4$. Then $E=2.33\sqrt{\frac{0.4\times(1 - 0.4)}{800}}\approx0.0446$. So the answer is A.

Step15: Solve for 10)

For $C = 0.90$, $\alpha=0.10$, $\alpha/2=0.05$, $z_{\alpha/2}=1.645$, $n = 1510$, $\hat{p}=0.35$. Then $E=1.645\sqrt{\frac{0.35\times(1 - 0.35)}{1510}}\approx0.0202$. So the answer is D.

Step16: Recall sample - size formula

When $\hat{p}$ and $\hat{q}$ are unknown, we use $\hat{p}=\hat{q}=0.5$ to get the maximum value of $\hat{p}\hat{q}$. The sample size formula is $n=\frac{z_{\alpha/2}^{2}\hat{p}\hat{q}}{E^{2}}$.

Step17: Solve for 11)

For $C = 0.93$, $\alpha=1 - 0.93 = 0.07$, $\alpha/2=0.035$, $z_{\alpha/2}\approx1.81$. Using $\hat{p}=\hat{q}=0.5$ and $E = 0.012$, $n=\frac{(1.81)^{2}\times0.5\times0.5}{(0.012)^{2}}\approx5688$. So the answer is B.

Answer:

1. C. 1.70
2. D. 1.62
3. B. 0.46 ± 0.08
4. D. 0.476 ± 0.044
5. C. 0.775
6. B. 0.019
7. D. 0.0474
8. C. 0.0184
9. A. 0.0446
10. D. 0.0202
11. B. 5688