# A random sample of 324 medical doctors showed that 164 had a

A random sample of 324 medical doctors showed that 164 had a solo practice.
a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.)
b) Find a 90% confidence interval for p. (Use 3 decimal places.)
lower limit
upper limit
c) Give a brief explanation of the meaning of the interval.
10% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.
90% of all confidence intervals would include the true proportion of physicians with solo practices.
10% of all confidence intervals would include the true proportion of physicians with solo practices.
90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.
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Jenny Bolton
Step 1
a) Let p represent the proportion of all medical doctors who have a solo practice.
The point estimate for p is,
$\stackrel{^}{p}=\frac{164}{324}$
$=0.506$
Step 2
b) 90% confidence interval for p:
From the given information, the confidence level is 90%.
Significance level is 10%.
From the standard normal table, the critical value corresponding with 10% significance level is 1.645.
$\text{lower limit}=\stackrel{^}{p}-{z}_{critical}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.506-\left(1.645×\sqrt{\frac{0.506\left(1-0.506\right)}{324}}\right)$
$=0.506-0.046$
$=0.460$
$\text{upper limit}=\stackrel{^}{p}+{z}_{critical}\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.506+\left(1.645×\sqrt{\frac{0.506\left(1-0.506\right)}{324}}\right)$
$=0.506+0.046$
$=0.552$
Step 3
c) Interpretation: 90% of all confidence intervals would include the true proportion of physicians with solo practices and 10% of all confidence intervals would not include the true proportion of physicians with solo practices. Correct option: 90% of all confidence intervals would include the true proportion of physicians with solo practices.
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Navreaiw

Step 1
a) Sample proportion $=0.494$
b) Sample size, $n=324$
Standard error, $SE=\sqrt{\frac{p\cap ×\left(1-p\cap \right)}{n}}$
$SE=\sqrt{\frac{0.494×\left(1-0.494\right\}\left\{324\right\}}{=}0.0278}$
Given Cl level is 90%, hence $\alpha =1-0.9=0.1$
$\frac{\alpha }{2}=\frac{0.1}{2}=0.05$
${Z}_{c}={Z}_{\frac{\alpha }{2}}=1.64$

c) Margin of Error, $ME=zc×SE$
$ME=1.64×0.0278$
$ME=0.046$