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Q.11.100

Expert-verifiedFound in: Page 476

Book edition
9th

Author(s)
Weiss, Neil

Pages
590 pages

ISBN
9780321989505

${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30;$

left-tailed test, $\alpha =0.10;80\%$ 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 $0.084$.

Given in the question that,

${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30;$

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

we need to determine the sample proportions.

The given values are,${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30,\alpha =0.10$, and $80\%$ confidence interval.

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

${\stackrel{~}{p}}_{1}=\frac{{x}_{1}}{{n}_{1}}$

Substitute ${x}_{1}=10,{n}_{1}=20$

${\stackrel{~}{p}}_{1}=\frac{10}{20}$

$=0.5$

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

${\stackrel{~}{p}}_{2}=\frac{{x}_{2}}{{n}_{2}}$

${\stackrel{~}{p}}_{2}=\frac{18}{30}$

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

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

Given in the question that,

${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30$;

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

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

The given values are, ${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30,\alpha =0.10$, 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.

Given in the question that,

${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30$

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

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

The given values are, ${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30,\alpha =0.10$, and $80\%$ confidence interval.

The formula for $z$is given by,

$z=\frac{{\stackrel{~}{p}}_{1}-{\stackrel{~}{p}}_{2}}{\sqrt{{\stackrel{~}{p}}_{p}\left(1-{\stackrel{~}{p}}_{p}\right)}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}$

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

${\stackrel{~}{p}}_{p}=\frac{{x}_{1}+{x}_{2}}{{n}_{1}+{n}_{2}}$

Substitute ${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30$

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

$=\frac{28}{50}$

$=0.56$

The value of$z$ is calculated as,

$z=\frac{{\stackrel{~}{p}}_{1}-{\stackrel{~}{p}}_{2}}{\sqrt{{\stackrel{~}{p}}_{p}\left(1-{\stackrel{~}{p}}_{p}\right)}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}$

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

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

$=-0.698$

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

${z}_{\alpha}=1.282$

${z}_{0.1}=1.282$

$z-1.282$ 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.

Given in the question that,

${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30$;

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

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

The given values are, ${x}_{1}=10,{n}_{1}=20,{x}_{2}=18,{n}_{2}=30,\alpha =0.10$, and $80\%$confidence interval.

For confidence level of $(1-\alpha )$ the confidence interval for ${p}_{1}-{p}_{2}$ are

$\left({\hat{p}}_{1}-{\hat{p}}_{2}\right)\pm {z}_{1/2}\times \sqrt{{\hat{p}}_{1}\left(1-{\hat{p}}_{1}\right)/{n}_{1}+{\hat{p}}_{2}\left(1-{\hat{p}}_{2}\right)/{n}_{2}}$

Calculate the value of $\alpha $,

$80=100(1-\alpha )$

$\alpha =0.2$

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

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

$\left({\stackrel{~}{p}}_{1}-{\stackrel{~}{p}}_{2}\right)\pm {z}_{\alpha /2}\xb7\sqrt{\frac{{\stackrel{~}{p}}_{1}\left(1-{\stackrel{~}{p}}_{1}\right)}{{n}_{1}+2}+\frac{{\stackrel{~}{p}}_{2}\left(1-{\stackrel{~}{p}}_{2}\right)}{{n}_{2}+2}}=(0.5-0.6)\pm 1.282$

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

$=-0.1\pm 0.184$

$=-0.284$ to $0.084$

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

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