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Q.11.100

Expert-verified

Elementary Statistics

Found in: Page 476

Book edition 9th

Author(s) Weiss, Neil

Pages 590 pages

ISBN 9780321989505

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Short Answer

$x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30;$
left-tailed test, $α = 0.10; 80 %$ confidence interval
a. Determine the sample proportions.
b. Decide whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d).
c. Use she two-proportions z-test to conduct the required hypothesis test.
d. Use the two-proportions z-interval procedure to find the specified confidence interval.

(a) The sample proportions are $0.5$ and $0.6$ .

(b) The two-proportion z-Procedure is appropriate

(c) The data does not provide sufficient evidence to reject the null hypothesis at the $10 %$ level of significance.

(d) The specified confidence interval is $- 0.284$ to $0.084$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a) Step 1: Given information

Given in the question that,

$x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30;$

left-tailed test, $α = 0.10; 80 %$ confidence interval

we need to determine the sample proportions.

Part(a) Step 2: Explanation

The given values are, $x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30, α = 0.10$ , and $80 %$ confidence interval.

The formula for ${\tilde{p}}_{1}$ is given by,

${\tilde{p}}_{1} = \frac{x_{1}}{n_{1}}$

Substitute $x_{1} = 10, n_{1} = 20$

${\tilde{p}}_{1} = \frac{10}{20}$

$= 0.5$

The formula for ${\tilde{p}}_{2}$ is given by,

${\tilde{p}}_{2} = \frac{x_{2}}{n_{2}}$

${\tilde{p}}_{2} = \frac{18}{30}$

role="math" localid="1651479624421" $= 0.6$

As a result the sample proportions are $0.5$ and $0.6$ .

Part (b) Step 1: Given information

Given in the question that,

$x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30$ ;

left-tailed test, $α = 0.10; 80 %$ confidence interval

we need to decide that whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d).

Part(b) Step 2: Explanation

The given values are, $x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30, α = 0.10$ , and $80 %$ confidence interval.

To begin, calculate $n_{1} - x_{1}$ and $n_{2} - x_{2}$ . After that, compare the outcome to 5. The two-proportion z-procedure technique is appropriate if it is more than or equal to 5.

The value of $n_{1} - x_{1}$ is calculated as,

$n_{1} - x_{1} = 20 - 10$

$= 10$

The value of $n_{2} - x_{2}$ is calculated as,

$n_{2} - x_{2} = 30 - 18$

$= 12$

The two-proportion z-procedure technique is appropriate because the values are more than $5$ . As a result, the two-proportion z-Procedure is appropriate.

Part (c) Step 1: Given information

Given in the question that,

$x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30$

left-tailed test, $α = 0.10; 80 %$ confidence interval

we need to use the two-proportions z-test to conduct the required hypothesis test.

Part (c) Step 2: Explanation

The given values are, $x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30, α = 0.10$ , and $80 %$ confidence interval.

The formula for $z$ is given by,

$z = \frac{{\tilde{p}}_{1} - {\tilde{p}}_{2}}{\sqrt{{\tilde{p}}_{p} (1 - {\tilde{p}}_{p})} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}$

The formula for ${\tilde{p}}_{p}$ is given by,

${\tilde{p}}_{p} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}}$

Substitute $x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30$

$= \frac{10 + 18}{20 + 30}$

$= \frac{28}{50}$

$= 0.56$

Part (c) Step 3: Value of z

The value of $z$ is calculated as,

$z = \frac{{\tilde{p}}_{1} - {\tilde{p}}_{2}}{\sqrt{{\tilde{p}}_{p} (1 - {\tilde{p}}_{p})} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}}$

$= \frac{0.5 - 0.6}{\sqrt{0.56 (1 - 0.56)} (\sqrt{\frac{1}{20} + \frac{1}{30}})}$

$= \frac{- 0.1}{0.143}$

$= - 0.698$

Perform the test at $10 %$ level of significance that is $α = 0.1$ from table-IV (at the bottom) the value of

$z_{α} = 1.282$

$z_{0.1} = 1.282$

$z - 1.282$ is the rejected region. As a result, the test static does not fall into the reject zone. As a result, the hypothesis $H_{o}$ is rejected, and the test findings at the $10 %$ level are not statistically significant.

As a result, the data does not provide sufficient evidence to reject the null hypothesis at the $10 %$ level of significance.

Part (d) Step 1: Given information

Given in the question that,

$x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30$ ;

left-tailed test, $α = 0.10; 80 %$ confidence interval

we need to find the specified confidence interval by using the two-proportions z-interval procedure

Part (d) Step 2: Explanation

The given values are, $x_{1} = 10, n_{1} = 20, x_{2} = 18, n_{2} = 30, α = 0.10$ , and $80 %$ confidence interval.

For confidence level of $(1 - α)$ the confidence interval for $p_{1} - p_{2}$ are

$({\hat{p}}_{1} - {\hat{p}}_{2}) \pm z_{1 / 2} \times \sqrt{{\hat{p}}_{1} (1 - {\hat{p}}_{1}) / n_{1} + {\hat{p}}_{2} (1 - {\hat{p}}_{2}) / n_{2}}$

Calculate the value of $α$ ,

$80 = 100 (1 - α)$

$α = 0.2$

The value of $z$ at $α / 2$ from the $z$ -score table is $1.282$ .

For the difference between the two-population proportion, the needed confidence interval is determined as,

$({\tilde{p}}_{1} - {\tilde{p}}_{2}) \pm z_{α / 2} \cdot \sqrt{\frac{{\tilde{p}}_{1} (1 - {\tilde{p}}_{1})}{n_{1} + 2} + \frac{{\tilde{p}}_{2} (1 - {\tilde{p}}_{2})}{n_{2} + 2}} = (0.5 - 0.6) \pm 1.282$

$. \sqrt{\frac{0.5 (1 - 0.5)}{20} + \frac{0.6 (1 - 0.6)}{30}}$

$= - 0.1 \pm 0.184$

$= - 0.284$ to $0.084$

As a result, the difference in adult-American percentages is $- 0.284$ to $0.084$ .

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